学习笔记之15道简单算法题

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15道简单算法题

http://www.cnblogs.com/hlxs/archive/2014/06/06/3772333.html

(●—●) | 剑指Offer_编程题_牛客网

http://www.nowcoder.com/ta/coding-interviews?page=1

source code

interview/学习笔记之15道简单算法题 at main · haoran119/interview (github.com)

1:合并排序，将两个已经排序的数组合并成一个数组，其中一个数组能容下两个数组的所有元素;

 1 #include <iostream>
 2 using namespace std;
 3 
 4 class Solution 
 5 {
 6 public:
 7     void MergeArray(int a[],int alen,int b[],int blen)
 8     {
 9          int len = alen + blen - 1;
10          alen --;
11          blen --;
12          
13          // 用归并排序的思想把两个数组合并 
14          while (alen >= 0 && blen >= 0)
15          {
16                if (a[alen] > b[blen])
17                   a[len --] = a[alen --];
18                else
19                   a[len --] = b[blen --];
20          }
21          
22          while (blen >= 0)
23                a[len --] = b[blen --];
24     }
25 };
26 
27 int main ()
28 {
29     Solution testSolution;
30     int arrA[] = {1, 2, 4, 6, 8, 0, 0, 0, 0, 0};
31     int arrB[] = {1, 3, 5, 7, 9};
32       
33     testSolution.MergeArray(arrA, 5, arrB, 5);
34     
35     for (int i = 0; i < 10; i ++)
36         cout << arrA[i] << endl;
37         
38     getchar();
39     
40     return 0;
41 }

View Code

2:合并两个单链表;

C++11最好用nullptr初始化指针得到空指针，同时尽量避免使用NULL。
- https://msdn.microsoft.com/zh-cn/library/4ex65770.aspx
复习struct，链表也不熟悉了，用当前节点还是next节点的时候出了几次错。
struct_百度百科
- http://baike.baidu.com/link?url=PxVQxXElF_rtYMqYhfNBzsjvTl9cG2PSKTqFZGV3IBAEHosImlkuaSMqCIxnShEqtDD4xA8gISTtP7rS7TENj_
struct (C++)
- https://msdn.microsoft.com/zh-cn/library/64973255.aspx

  1 #include <iostream>
  2 using namespace std;
  3 
  4 struct ListNode 
  5 {
  6        int val;
  7        ListNode *next;
  8        ListNode(int x) : val(x), next(NULL) {}
  9 };
 10  
 11 class Solution 
 12 {
 13 public:
 14     ListNode* MergeList(ListNode* aListNode, ListNode* bListNode)
 15     {
 16          // NULL case
 17          if (aListNode == NULL)
 18             return bListNode;
 19          if (bListNode == NULL)
 20             return aListNode;
 21 
 22          ListNode* resHead = NULL;         
 23 
 24          if (aListNode->val < bListNode->val) {
 25             resHead = aListNode;
 26             aListNode = aListNode->next;
 27          } else {
 28             resHead = bListNode;
 29             bListNode = bListNode->next;
 30          }
 31          
 32          // Reserve the head point         
 33          ListNode* tmpHead = resHead;
 34 
 35 //         PrintListNode(tmpHead);
 36          
 37          // Merge two LinkList
 38          // Remember to check if the current node is NULL
 39          // assign ListNode to next resHead->next
 40          // move resHead to the next ptr
 41          while (aListNode != NULL && bListNode != NULL)
 42          {
 43                if (aListNode->val < bListNode->val) {
 44                   resHead->next = aListNode;
 45                   aListNode = aListNode->next;
 46                } else {
 47                   resHead->next= bListNode;
 48                   bListNode = bListNode->next;
 49                }
 50                resHead = resHead->next;
 51          }
 52 
 53 //         PrintListNode(tmpHead);
 54 
 55          if (aListNode != NULL)
 56             resHead->next= aListNode;
 57             
 58          if (bListNode != NULL)
 59             resHead->next= bListNode;
 60          
 61 //         PrintListNode(tmpHead);
 62 
 63          return tmpHead;
 64     }
 65     
 66     void PrintListNode(ListNode* inListNode)
 67     {
 68          ListNode* tmpLN = inListNode;
 69          
 70          while (tmpLN != NULL)
 71          {
 72                cout << tmpLN->val << endl;
 73                tmpLN = tmpLN->next;
 74          }         
 75          cout << endl;
 76     }
 77 };
 78 
 79 int main ()
 80 {
 81     Solution testSolution;
 82     ListNode* aHead = new ListNode(1);
 83     ListNode* cur = aHead;
 84     
 85     for (int i = 2; i < 10; i += 2)
 86     {
 87         ListNode* tmpLN = new ListNode(i);
 88         cur->next = tmpLN;
 89         cur = tmpLN;        
 90     }
 91     
 92 //    testSolution.PrintListNode(aHead);
 93     
 94     ListNode* bHead = new ListNode(1);
 95     cur = bHead;
 96     
 97     for (int i = 3; i < 10; i += 2)
 98     {
 99         ListNode* tmpLN = new ListNode(i);
100         cur->next = tmpLN;
101         cur = tmpLN;
102     }
103       
104 //    testSolution.PrintListNode(bHead);
105 
106     ListNode* head = testSolution.MergeList(aHead, bHead);
107 
108 //    testSolution.PrintListNode(head);
109 
110     while (head != NULL)
111     {
112           cout << head->val << endl;
113           head = head->next;
114     }
115         
116     getchar();
117     
118     return 0;
119 }

View Code

3:倒序打印一个单链表;

递归
链表的题目还得多练，又弄错了打印head还是cur节点。。。

 1 #include <iostream>
 2 using namespace std;
 3 
 4 struct ListNode 
 5 {
 6        int val;
 7        ListNode *next;
 8        ListNode(int x) : val(x), next(NULL) {}
 9 };
10  
11 class Solution 
12 {
13 public:
14     void PrintListNodeReversely(ListNode* inListNode)
15     {
16          if (inListNode != NULL)
17          {
18             PrintListNodeReversely(inListNode->next);
19             cout << inListNode->val << endl;
20          }         
21     }
22 };
23 
24 int main ()
25 {
26     Solution testSolution;
27     ListNode* head = new ListNode(0);
28     ListNode* cur = head;
29     
30     for (int i = 1; i < 10; i ++)
31     {
32         ListNode* tmpLN = new ListNode(i);
33         cur->next = tmpLN;
34         cur = cur->next; // cur->next == tmpLN        
35     }
36 
37     // Print head node        
38     testSolution.PrintListNodeReversely(head);
39         
40     getchar();
41     
42     return 0;
43 }

View Code

4:给定一个单链表的头指针和一个指定节点的指针，在O(1)时间删除该节点;

http://www.cnblogs.com/pegasus923/archive/2010/10/04/1841891.html
重温链表插入删除操作，注意原文代码删除头结点处有问题，函数对头指针只是值传递，如果要删除需要传入头指针的指针。

  1 #include <iostream>
  2 using namespace std;
  3 
  4 struct ListNode 
  5 {
  6        int val;
  7        ListNode *next;
  8        ListNode(int x) : val(x), next(NULL) {}
  9 };
 10  
 11 class Solution 
 12 {
 13 public:
 14     void PrintListNode(ListNode* inListNode)
 15     {
 16          if (inListNode != NULL) {
 17             cout << inListNode->val << endl;
 18             PrintListNode(inListNode->next);
 19          } else
 20             cout << endl;
 21     }
 22 
 23     void Test(ListNode* inListNode)
 24     {
 25          delete inListNode;
 26          inListNode = NULL;
 27          cout << inListNode << endl;
 28 
 29          PrintListNode(inListNode);
 30     }
 31     
 32     void DeleteNode(ListNode* pInHead, ListNode* pToBeDeleted)
 33     {
 34          cout << "Entry void DeleteNode(ListNode* pInHead, ListNode* pToBeDeleted)\n" << endl;
 35 
 36          // Check NULL always
 37          if (pInHead == NULL || pToBeDeleted == NULL)
 38             return;
 39          
 40 //         PrintListNode(pInHead);
 41          
 42          // Delete non-tail node including head node
 43          if (pToBeDeleted->next != NULL) {
 44             ListNode* pNext = pToBeDeleted->next;
 45             pToBeDeleted->val = pNext->val;
 46             pToBeDeleted->next = pNext->next;
 47             
 48             delete pNext;
 49             pNext = NULL;
 50          } else { // Delete tail
 51             ListNode* pPre = pInHead;
 52          
 53             while (pPre->next != pToBeDeleted && pPre != NULL) 
 54                   pPre = pPre->next;
 55          
 56             if (pPre == NULL)
 57                return;
 58          
 59             pPre->next = NULL;
 60             delete pToBeDeleted;
 61             pToBeDeleted = NULL;
 62          }                  
 63     }
 64 };
 65 
 66 int main ()
 67 {
 68     Solution testSolution;
 69     int count = 5;
 70  
 71     for (int k = 0; k <= count; k ++)
 72     {
 73         ListNode* pHead = NULL;
 74         ListNode* pCur = NULL;
 75         ListNode* pDel = NULL;
 76         
 77         for (int i = 0; i < count; i ++)
 78         {
 79             ListNode* pTemp = new ListNode(i);
 80             
 81             if (i == 0)
 82                 pHead = pCur = pTemp;
 83             else {
 84                 pCur->next = pTemp;
 85                 pCur = pCur->next; // pCur->next == pTemp
 86             }
 87             
 88             if (i == k) 
 89                pDel = pTemp;
 90         }
 91            
 92         cout << "pHead = " << pHead << endl;
 93         if (pHead != NULL)
 94            cout << "pHead->val = " << pHead->val << endl;
 95         cout << "pDel = " << pDel << endl;
 96         if (pDel != NULL)
 97            cout << "pDel->val = " << pDel->val << endl;
 98         cout << ((pHead == pDel) ? "pHead == pDel" : "pHead != pDel") << endl;
 99         cout << endl;
100 
101        // Print Linkedlist
102         testSolution.PrintListNode(pHead);        
103          
104         testSolution.DeleteNode(pHead, pDel);
105 //        testSolution.Test(pHead);
106 
107 /*
108         delete pDel;
109         pDel = NULL;
110         pHead = NULL;
111 */        
112         cout << "pHead = " << pHead << endl;
113         if (pHead != NULL)
114            cout << "pHead->val = " << pHead->val << endl;
115         cout << "pDel = " << pDel << endl;
116         if (pDel != NULL)
117            cout << "pDel->val = " << pDel->val << endl;
118         cout << endl;
119 
120         // Print Linkedlist
121         testSolution.PrintListNode(pHead);        
122     }   
123         
124     getchar();
125     
126     return 0;
127 }

View Code

5:找到链表倒数第K个节点; K=0时，返回链表尾元素。

两个指针同时往后走，之间相隔K个元素。一个到达尾部时，另一个正好为倒数第K个。
注意边界情况，头指针为NULL，K为0，K大于链表长度。

  1 #include <iostream>
  2 using namespace std;
  3 
  4 struct ListNode 
  5 {
  6        int val;
  7        ListNode *next;
  8        ListNode(int x) : val(x), next(NULL) {}
  9 };
 10  
 11 class Solution 
 12 {
 13 public:
 14     void PrintListNode(ListNode* inListNode)
 15     {
 16          if (inListNode != NULL) {
 17             cout << inListNode->val << endl;
 18             PrintListNode(inListNode->next);
 19          } else
 20             cout << endl;
 21     }
 22     
 23     ListNode *findMToLastElement(ListNode* pInHead, int m)
 24     {
 25          cout << "Entry ListNode *findMToLastElement(ListNode* pInHead, int m)\n" << endl;
 26          
 27          ListNode *pCur = NULL, *pMPre = NULL;
 28          
 29          // Corner case
 30          if (pInHead == NULL) return NULL;
 31          
 32          pCur = pMPre = pInHead;
 33 
 34          // Move pCur forward m elements
 35          for (int i = 0; i < m; i ++)
 36          {
 37              if (pCur->next != NULL)
 38                 pCur = pCur->next;
 39              else
 40                 return NULL;
 41          }
 42 
 43          // Move forward together until pCur reach the tail
 44          while (pCur->next != NULL)
 45          {
 46                pCur = pCur->next;
 47                pMPre = pMPre->next;               
 48          }
 49          
 50          // Now pMPre points to the M to last element
 51          return pMPre;
 52     }
 53 };
 54 
 55 int main ()
 56 {
 57     Solution testSolution;
 58     int count = 10;
 59     
 60     ListNode *pHead = NULL;
 61     ListNode *pCur = NULL;
 62     
 63     for (int i = 0; i < count; i ++)
 64     {
 65         ListNode *pTemp = new ListNode(i);
 66         
 67         if (i == 0)
 68             pHead = pCur = pTemp;
 69         else {
 70             pCur->next = pTemp;
 71             pCur = pCur->next; // pCur->next == pTemp
 72         }
 73     }
 74     
 75     // Print Head       
 76     cout << "pHead = " << pHead << endl;
 77     if (pHead != NULL)
 78        cout << "pHead->val = " << pHead->val << endl;
 79     cout << endl;
 80     
 81     // Print Linkedlist
 82     cout << "PrintListNode(pHead)" << endl;
 83     testSolution.PrintListNode(pHead);        
 84 
 85     cout << "Test findMToLastElement(pHead, 0)\n" << endl;     
 86     ListNode *pMToLastElement = testSolution.findMToLastElement(pHead, 0);
 87     
 88     // Print Head       
 89     cout << "pHead = " << pHead << endl;
 90     if (pHead != NULL)
 91        cout << "pHead->val = " << pHead->val << endl;
 92     cout << endl;
 93     
 94     // Print M to last element
 95     cout << "pMToLastElement = " << pMToLastElement << endl;
 96     if (pMToLastElement != NULL)
 97        cout << "pMToLastElement->val = " << pMToLastElement->val << endl;
 98     cout << endl;
 99     
100     getchar();
101     
102     return 0;
103 }

View Code

6:反转单链表;

注意保留反转时，先保留next指针地址，反转之后再赋值给current指针往后走，否则丢了next指针。

  1 #include <iostream>
  2 using namespace std;
  3 
  4 struct ListNode 
  5 {
  6        int val;
  7        ListNode *next;
  8        ListNode(int x = 0) : val(x), next(NULL) {}
  9 };
 10  
 11 class Solution 
 12 {
 13 public:
 14     void PrintList(ListNode *inListNode)
 15     {
 16          if (inListNode != NULL) {
 17             cout << inListNode->val << endl;
 18             PrintList(inListNode->next);
 19          } else
 20             cout << "NULL\n" << endl;
 21     }
 22     
 23     // Assign in ptr of the head ptr
 24     void DeleteList(ListNode **pInHead)
 25     {
 26          cout << "Entry void DeleteList(ListNode **pInHead)\n" << endl;
 27          
 28          ListNode *pDel = *pInHead;
 29          
 30          while (pDel != NULL)
 31          {
 32                // Reserve the address of the next node
 33                ListNode *pNext = pDel->next;              
 34                // Delete the current node
 35                delete pDel;
 36                pDel = NULL;
 37                // Move to the next node
 38                pDel = pNext;
 39          }
 40 
 41          // Set head to NULL         
 42          *pInHead = NULL;
 43     }
 44     
 45     ListNode* ReverseList(ListNode* pHead)
 46     {
 47          cout << "Entry ListNode* ReverseList(ListNode* pHead)\n" << endl;
 48 
 49          // Corner case
 50          if (pHead == NULL)       return NULL;
 51          
 52          ListNode *pPre = NULL, *pRHead = NULL, *pCur = pHead;
 53          
 54          while (pCur != NULL)
 55          {
 56                // Reserve the next node
 57                ListNode *pNext = pCur->next;               
 58                // Reserve the address of the last node of original list
 59                if (pNext == NULL) pRHead = pCur;
 60                // Insert current node
 61                pCur->next = pPre;
 62                // Move the previous ptr
 63                pPre = pCur;
 64                // Assign the next node back
 65                pCur = pNext;
 66          }
 67          
 68          return pRHead;
 69     }
 70 };
 71 
 72 int main ()
 73 {
 74     Solution testSolution;
 75     int count = 10;
 76     
 77     ListNode *pMHead = NULL;
 78     ListNode *pMCur = NULL;
 79     
 80     for (int i = 0; i < count; i ++)
 81     {
 82         ListNode *pTemp = new ListNode(i);
 83         
 84         if (i == 0)
 85             pMHead = pMCur = pTemp;
 86         else {
 87             pMCur->next = pTemp;
 88             pMCur = pMCur->next; // pCur->next == pTemp
 89         }
 90     }
 91     
 92     // Print Head       
 93     cout << "pMHead = " << pMHead << endl;
 94     if (pMHead != NULL)
 95        cout << "pMHead->val = " << pMHead->val << endl;
 96     cout << endl;
 97     
 98     // Print Linkedlist
 99     cout << "PrintList(pMHead)" << endl;
100     testSolution.PrintList(pMHead);        
101 
102     cout << "Test ReverseList(pMHead)\n" << endl;     
103     ListNode *pMRHead = testSolution.ReverseList(pMHead);
104     
105     // Print Reversed Head       
106     cout << "pMRHead = " << pMRHead << endl;
107     if (pMRHead != NULL)
108        cout << "pMRHead->val = " << pMRHead->val << endl;
109     cout << endl;
110     
111     // Print Linkedlist
112     cout << "PrintList(pMRHead)" << endl;
113     testSolution.PrintList(pMRHead);        
114 
115     // Print Head       
116     cout << "pMHead = " << pMHead << endl;
117     if (pMHead != NULL)
118        cout << "pMHead->val = " << pMHead->val << endl;
119     cout << endl;
120     
121     // Print Linkedlist
122     cout << "PrintList(pMHead)" << endl;
123     testSolution.PrintList(pMHead);        
124 
125     cout << "DeleteList(&pMRHead)" << endl;
126     testSolution.DeleteList(&pMRHead);
127     
128     // Print Reversed Head       
129     cout << "pMRHead = " << pMRHead << endl;
130     if (pMRHead != NULL)
131        cout << "pMRHead->val = " << pMRHead->val << endl;
132     cout << endl;
133     
134     // Print Linkedlist
135     cout << "PrintList(pMRHead)" << endl;
136     testSolution.PrintList(pMRHead);        
137 
138     // Print Head       
139     cout << "pMHead = " << pMHead << endl; // It is not NULL, the space is free'ed?
140     if (pMHead != NULL)
141        cout << "pMHead->val = " << pMHead->val << endl;
142     cout << endl;
143     
144     // Print Linkedlist
145 //    cout << "PrintList(pMHead)" << endl;
146 //    testSolution.PrintList(pMHead);        
147 
148     getchar();
149     
150     return 0;
151 }

View Code

7:通过两个栈实现一个队列;

stack 类
- https://msdn.microsoft.com/zh-cn/library/56fa1zk5.aspx
stack - C++ Reference
- 　　http://www.cplusplus.com/reference/stack/stack/
Stack_百度百科
- http://baike.baidu.com/link?url=umZEKUINQtIeQve3VZ9u7eJSI2IvARxNaxZus-XwejEcV2QSUjgD-qNQS-efMtjl_96xGF_LXYyUuGO0-ewAzq#5

 1 #include <iostream>
 2 #include <stack>
 3 using namespace std;
 4  
 5 class Solution 
 6 {
 7 public:
 8        void push(int node) 
 9        {
10             stack1.push(node);
11        }
12        
13        int pop() 
14        {
15             int result = 0;
16             
17             if (stack2.empty())
18             {
19                 while (!stack1.empty())
20                 {
21                       stack2.push(stack1.top());
22                       stack1.pop();
23                 }                  
24             }    
25             
26             result = stack2.top();
27             stack2.pop();
28             
29             return result;
30        }
31        
32 private:
33     stack<int> stack1;
34     stack<int> stack2;
35 };
36 
37 int main ()
38 {
39     Solution testSolution;
40     int count = 10;
41 
42     // Case 1    
43     cout << "CASE 1\n" << endl;
44     
45     for (int i = 0; i < count; i ++)
46     {
47         testSolution.push(i);
48     }
49     
50     for (int i = 0; i < count; i ++)
51     {
52         cout << testSolution.pop() << endl;    
53     }
54 
55     // Case 2
56     cout << "\nCASE 2\n" << endl;
57 
58     for (int i = 0; i < count; i ++)
59     {
60         testSolution.push(i);
61         cout << testSolution.pop() << endl;    
62     }
63 
64     // Case 3
65     cout << "\nCASE 3\n" << endl;
66 
67     for (int i = 0; i < count / 2; i ++)
68     {
69         testSolution.push(i);
70     }
71     
72     cout << testSolution.pop() << "\n" << endl;    
73 
74     for (int i = count / 2; i < count; i ++)
75     {
76         testSolution.push(i);
77     }
78     
79     for (int i = 0; i < count - 1; i ++)
80     {
81         cout << testSolution.pop() << endl;        
82     }
83 
84     getchar();
85     
86     return 0;
87 }

View Code

8:二分查找;

用vector实现递归和非递归二分查找。刚开始没弄清元素个数，初始化1个元素，再push_back n-1个元素，这时vector里应该是n个元素。
iterator作为入参传递有问题，先用int代替了，以后再研究
用distance来计算找到的当前元素是第几个元素
vector的二分查找算法 - 轻典 - 博客园
　　http://www.cnblogs.com/tianyajuanke/archive/2012/05/23/2515117.html
distance - C++ Reference
　　http://www.cplusplus.com/reference/iterator/distance/
distance
　　https://msdn.microsoft.com/zh-cn/library/k1kfxk7e.aspx

 1 #include <iostream>
 2 #include <vector>
 3 #include <iterator>
 4 using namespace std;
 5  
 6 class Solution 
 7 {
 8 public:
 9     int binarySearch(vector<int> array, int target)
10     {
11         vector<int>::iterator     vecBIT = array.begin();
12         vector<int>::iterator     vecEIT = array.end();
13         vector<int>::iterator     vecIT = vecBIT + (vecEIT - vecBIT) / 2;
14         
15         while (vecBIT <= vecEIT)  
16         {
17               if (*vecIT == target)
18                   return distance(array.begin(), vecIT); // Return distance between iterators
19               else if (*vecIT < target)
20               {                   
21                   vecBIT = vecIT + 1;
22                   vecIT = vecBIT + (vecEIT - vecBIT) / 2;
23               } 
24               else
25               {
26                   vecEIT = vecIT - 1;
27                   vecIT = vecBIT + (vecEIT - vecBIT) / 2;
28               }
29         }
30         
31         if (vecBIT > vecEIT) return -1;
32     }
33 
34     int binarySearchRecursion(vector<int> array, int vecBIT, int vecEIT, int target)
35     {       
36 //        cout << "Entry int binarySearchRecursion(vector<int> array, vector<int>::iterator vecBIT, vector<int>::iterator vecEIT, int target)\n" << endl;
37         
38         if (vecBIT <= vecEIT)  
39         {
40             int    vecIT = vecBIT + (vecEIT - vecBIT) / 2;
41             
42             if (array[vecIT] == target)
43               return vecIT;
44             else if (array[vecIT] < target)
45               return binarySearchRecursion(array, vecIT + 1, vecEIT, target);
46             else
47               return binarySearchRecursion(array, vecBIT, vecIT - 1, target);
48         } 
49         else
50             return -1;
51     }
52 };
53 
54 int main ()
55 {
56     Solution testSolution;
57     int n = 3;
58     vector<int> searchArray(1); // initialize vector of n elements with 0
59     
60     // push_back another n - 1 elements
61     for (int i = 1; i < n; i ++)
62         searchArray.push_back(i);
63         
64     // n element in vector now
65     for (vector<int>::const_iterator vecIT = searchArray.begin(); vecIT < searchArray.end(); vecIT ++)
66         cout << *vecIT << endl;
67     cout << endl;
68     
69     cout << distance(searchArray.begin(), searchArray.end()) << "\n" << endl;
70     
71     for (int i = 0; i <= n; i ++)
72     {
73         cout << testSolution.binarySearch(searchArray, i) << endl;
74         cout << testSolution.binarySearchRecursion(searchArray, 0, n, i) << endl;
75     }
76             
77     getchar();
78     
79     return 0;
80 }

View Code

9:快速排序;

注意对vector入参要用引用，不能用形参，否则对vector元素排序不起作用
vector有序情况下，快排时间复杂度最高退化为O(N^2)
输出vector也可以使用范围for语句
- for (auto vecCIT : vecArray) cout << vecCIT << endl;

 1 #include <iostream>
 2 #include <vector>
 3 using namespace std;
 4  
 5 class Solution 
 6 {
 7 public:
 8     // Pass in reference of vector parameter
 9     void quickSort(vector<int> &a, int inLeft, int inRight)
10     {       
11         cout << "Entry void qSort(vector<int> array, int left, int right)\n" << endl;
12         
13         if (inLeft >= inRight) 
14            return;
15         
16         int i = inLeft, j = inRight, temp = a[inLeft];
17 
18         while (i < j)
19         {
20               // Scan from right to left
21               while ((i < j) && (a[j] >= temp)) 
22                     j --;
23               if (i < j) 
24                  a[i ++] = a[j];
25                  
26               // Scan from left to right
27               while ((i < j) && (a[i] <= temp)) 
28                     i ++;
29               if (i < j)
30                  a[j --] = a[i];                         
31         }
32         // Base value temp is on the right position
33         a[i] = temp;
34         
35         // Divide and conquer
36         quickSort(a, inLeft, i - 1);
37         quickSort(a, i + 1, inRight);        
38     }
39 };
40 
41 int main ()
42 {
43     Solution testSolution;
44     int array[] = {0, 1, 5, 9, 0, 2, 6, 7, 8, 3, 4};
45 //    int array[] = {0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
46 //    int array[] = {9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 0};
47 //    int array[] = {9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9};
48     vector<int> vecArray(array, array + 11);
49     
50     for (vector<int>::const_iterator vecCIT = vecArray.begin(); vecCIT != vecArray.end(); vecCIT ++)
51         cout << *vecCIT << endl;
52     cout << endl;
53 
54     testSolution.quickSort(vecArray, 0, vecArray.size() - 1);
55         
56     for (vector<int>::const_iterator vecCIT = vecArray.begin(); vecCIT != vecArray.end(); vecCIT ++)
57         cout << *vecCIT << endl;
58     cout << endl;
59             
60     getchar();
61     
62     return 0;
63 }

View Code

10:获得一个int型的数中二进制中1的个数; 输入一个整数，输出该数二进制表示中1的个数。其中负数用补码表示。

位运算
如果不是负数的话，可以一直右移获取1的个数，f(n) = f(n / 2) + n % 2。跟LeetCode 338 Counting Bits一样。
- 　　http://www.cnblogs.com/pegasus923/p/5499737.html
负数用补码表示。所以只能用n = n & (n - 1)，用n跟n-1做&运算，把二进制中的1从后往前去掉，直到没有。注意2的幂的特征(n & (n - 1)) == 0，e.g. 0x1000 & 0x0111。
补码_百度百科
- 　　http://baike.baidu.com/link?url=ITDahaDRGLVt2jyxtz0t3a7DbXYFBk1kVZMxzX7SsmxRnTpPBYKHCBYM2zFs1t3s6FrDlgFoYqtTZnnbWClt6a

 1 #include <iostream>
 2 using namespace std;
 3  
 4 class Solution 
 5 {
 6 public:
 7     int  NumberOf1(int n)
 8     {       
 9         cout << "Entry int  NumberOf1(int n)\n" << endl;
10 
11         if (n == 0) return 0;
12         
13         int count = 0;
14         
15         while (n != 0)
16         {
17               n = n & (n - 1);
18               count ++;
19         }
20         
21         return count;
22     }
23 };
24 
25 int main ()
26 {
27     Solution testSolution;
28     int n = 10;
29         
30     for (int i = 0; i < n; i ++)
31         cout << testSolution.NumberOf1(i) << endl;
32         
33     getchar();
34     
35     return 0;
36 }

View Code

11:输入一个数组，实现一个函数，让所有奇数都在偶数前面; 并保证奇数和奇数，偶数和偶数之间的相对位置不变。

跟排序思想差不多。原文的思想类似快排，两个指针一个从头往后找偶数，另一个从尾往前找奇数，找到就交换，直到两个指针相遇。但这个是不稳定的，所以奇偶数之间的相对位置会变。可以用归并，冒泡排序思想，这些都是稳定的。
另一种解法就是借助辅助vector存储偶数，扫描一遍把偶数挑出来并从原vector删除，之后再push_back回去。
注意vector.end()并不是最后一个元素，而是vector结束符。要找最后一个元素的下标是vector.end()-1。

 1 #include <iostream>
 2 #include <vector>
 3 using namespace std;
 4  
 5 class Solution 
 6 {
 7 public:
 8     void reOrderArray(vector<int> &array)
 9     {       
10         cout << "Entry void reOrderArray(vector<int> &array)\n" << endl;
11         
12         vector<int> vecEven; // Store even
13         
14         for (vector<int>::iterator vecIT = array.begin(); vecIT != array.end(); )
15             if ((*vecIT % 2) == 0)
16             {
17                          vecEven.push_back(*vecIT);
18                          vecIT = array.erase(vecIT);
19             }
20             else
21                 vecIT ++;                         
22                 
23         for (vector<int>::const_iterator vecCIT = vecEven.begin(); vecCIT != vecEven.end(); vecCIT ++)
24             array.push_back(*vecCIT);            
25     }
26 };
27 
28 int main ()
29 {
30     Solution testSolution;
31     int array[] = {0, 1, 5, 9, 2, 6, 7, 8, 3, 4};
32     vector<int> vecArray(array, array + 10);
33         
34     for (vector<int>::const_iterator vecCIT = vecArray.begin(); vecCIT != vecArray.end(); vecCIT ++)
35         cout << *vecCIT << endl;
36     cout << endl;
37 
38     testSolution.reOrderArray(vecArray);
39         
40     for (vector<int>::const_iterator vecCIT = vecArray.begin(); vecCIT != vecArray.end(); vecCIT ++)
41         cout << *vecCIT << endl;
42     cout << endl;
43         
44     getchar();
45     
46     return 0;
47 }

View Code

12:判断一个字符串是否是另一个字符串的子串;

简单的暴力匹配。关键是熟悉string函数，不出BUG。
kmp算法_百度百科
- 　　http://baike.baidu.com/view/659777.htm
字符串匹配的KMP算法_知识库_博客园
- 　　http://kb.cnblogs.com/page/176818/
字符串匹配的Boyer-Moore算法_知识库_博客园
- 　　http://kb.cnblogs.com/page/176945/
KMP算法详解
- 　　http://www.matrix67.com/blog/archives/115

 1 #include <iostream>
 2 #include <string>
 3 using namespace std;
 4 
 5 class Solution {
 6 public:
 7     int substr(const string source, const string sub)
 8     {
 9         // Corner case
10         if (sub.empty()) return 0;
11         if (source.empty()) return -1;
12                 
13         int n = source.length() - sub.length() + 1;
14 
15         // better use < rather than <=
16         for (int i = 0; i < n; i ++)
17         {
18               bool flag = true;
19               
20               for (int j = 0; j < sub.length(); j ++)
21               {
22                   if (source.at(i + j) != sub.at(j))
23                   {
24                     flag = false;
25                     break;
26                   }
27               }
28               
29               if (flag) return i;
30         }
31         
32         return -1;
33     }    
34 };
35 
36 int main ()
37 {
38     Solution testSolution;
39     string s1 = "mississippi";
40     string s2 = "issi";
41     
42     cout << testSolution.substr(s1, s2) << endl;
43     
44     getchar();
45     
46     return 0;
47 }

View Code

13:把一个int型数组中的数字拼成一个串，这个串代表的数字最小;

把数组转成字符串然后进行升序排序，定义如果AB < BA，那么A < B。排序完成，再把数组组合就成了最小串了。
学习了sort的比较函数的写法。注意类中自定义的compare需要定义成static，否则sort函数编译出错。
学习了int跟string转换的几种方法，to_string方便只是是C++11的。
sort - C++ Reference
- 　　http://www.cplusplus.com/reference/algorithm/sort/?kw=sort
sort
- 　　https://msdn.microsoft.com/zh-cn/library/ecdecxh1(v=vs.110).aspx
STL 中sort、qsort 的用法 - rattles的专栏 - 博客频道 - CSDN.NET
- 　　http://blog.csdn.net/rattles/article/details/5510919
C++中Static作用和使用方法
- 　　http://blog.csdn.net/artechtor/article/details/2312766
sprintf - C++ Reference
- 　　http://www.cplusplus.com/reference/cstdio/sprintf/
sprintf、_sprintf_l、swprintf、_swprintf_l、__swprintf_l
- 　　https://msdn.microsoft.com/zh-cn/library/ybk95axf.aspx
to_string - C++ Reference
- 　　http://www.cplusplus.com/reference/string/to_string/
C++ int与string的转化 - Andy Niu - 博客园
- 　　http://www.cnblogs.com/nzbbody/p/3504199.html

 1 #include <iostream>
 2 #include <vector>
 3 #include <cstdio> // sprintf
 4 #include <algorithm>  // sort
 5 using namespace std;
 6  
 7 class Solution 
 8 {
 9 public:
10     // Need to define static
11     static bool compare(const string &sA, const string &sB)
12     {
13            string sAB = sA + sB;
14            string sBA = sB + sA;
15 
16          return sAB < sBA;
17     }
18        
19     string PrintMinNumber(vector<int> numbers)
20     {       
21         cout << "Entry string PrintMinNumber(vector<int> numbers)\n" << endl;
22 
23         string result = "";
24 
25         // Corner case
26         if (numbers.size() == 0) return result;
27         
28         vector<string> sNumbers;
29         char *sTemp = new char[10]; // Remember to initialize ptr
30         
31         for (vector<int>::const_iterator vecCIT = numbers.begin(); vecCIT != numbers.end(); vecCIT ++)
32         {
33 //         string sTemp = to_string(*vecCIT); // C++11
34             sprintf(sTemp, "%d", *vecCIT);
35             sNumbers.push_back(sTemp);
36         }
37 
38         // Use self-defined compare function
39         sort(sNumbers.begin(), sNumbers.end(), compare);
40 
41         for (vector<string>::const_iterator vecCIT = sNumbers.begin(); vecCIT != sNumbers.end(); vecCIT ++)
42             result += *vecCIT;
43 
44         return result;
45     }
46 };
47 
48 int main ()
49 {
50     Solution testSolution;
51     int array[] = {3, 32, 321};
52     vector<int> vecArray(array, array + 3);
53 //    vector<int> vecArray;
54         
55     for (vector<int>::const_iterator vecCIT = vecArray.begin(); vecCIT != vecArray.end(); vecCIT ++)
56         cout << *vecCIT << endl;
57     cout << endl;
58 
59     cout << testSolution.PrintMinNumber(vecArray) << endl;
60                 
61     getchar();
62     
63     return 0;
64 }

View Code

14:输入一颗二叉树，输出它的镜像（每个节点的左右子节点交换位置）;

交换左右子节点（注意是节点，不只是值），然后递归处理左右子树。
复习了建树过程。
swap - C++ Reference
- 　　http://www.cplusplus.com/reference/utility/swap/?kw=swap

 1 #include <iostream>
 2 #include <vector>
 3 using namespace std;
 4 
 5 struct TreeNode {
 6     int val;
 7     struct TreeNode *left;
 8     struct TreeNode *right;
 9     TreeNode(int x) :
10             val(x), left(NULL), right(NULL) {
11     }
12 };
13  
14 class Solution 
15 {
16 public:
17     TreeNode* CreateTree(const vector<int> &vec, const int &pos)
18     {
19 //              cout << "Entry TreeNode* CreateTree(const vector<int> &vec, const int &pos)\n" << endl;
20               
21 //              cout << "pos = " << pos << endl;
22 //              cout << "vec.size() = " << vec.size() << endl;
23             
24               // Pay attention to >=, or else infinite loop    
25               if (pos >= vec.size()) return NULL;
26               
27 //              cout << "vec[pos] = " << vec[pos] << endl;
28 
29               TreeNode *pCur = new TreeNode(vec[pos]);
30 //              cout << "pCur->val = " << pCur->val<< endl;
31               pCur->left = CreateTree(vec, 2 * pos);
32               pCur->right = CreateTree(vec, 2 * pos + 1);              
33               
34               return pCur;
35     }
36     
37     void PrintTree(TreeNode *pTreeNode)
38     {
39          if (pTreeNode != NULL) {
40             cout << pTreeNode->val << endl;
41             PrintTree(pTreeNode->left);
42             PrintTree(pTreeNode->right);
43          } 
44     }
45     
46     void Mirror(TreeNode *pRoot) 
47     {
48 //         cout << "Entry void Mirror(TreeNode *pRoot) " << endl;
49          
50          // Corner case
51          if ((pRoot == NULL) || ((pRoot->left == NULL) && (pRoot->right == NULL))) return;
52          
53          // swap two ptr rathet than value
54          swap(pRoot->left, pRoot->right);
55          
56          Mirror(pRoot->left);
57          Mirror(pRoot->right);                  
58     }
59 };
60 
61 int main ()
62 {
63     Solution testSolution;
64     int count = 8;
65     int array[] = {0, 8, 6, 10, 5, 7, 9, 11};
66     vector<int> vecArray(array, array + count);
67 
68     TreeNode *pMRoot = testSolution.CreateTree(vecArray, 1);
69 /*
70     cout << root << endl;
71     cout << root->val << endl;
72 */    
73     testSolution.PrintTree(pMRoot);
74     cout << endl;
75 
76     testSolution.Mirror(pMRoot);
77 
78     testSolution.PrintTree(pMRoot);
79     cout << endl;
80     
81     getchar();
82     
83     return 0;
84 }

View Code

15:输入两个链表，找到它们第一个公共节点;

若有公共节点，则从公共节点之后的所有节点都是公共的。I.e. 短链表是长链表的子串。基于这个思想，长链表里的公共节点只可能出现在倒数K个节点里（K为短链表长度）。长链表先走N-K步，长度相同时再开始往后挨个找。

  1 #include <iostream>
  2 using namespace std;
  3 
  4 struct ListNode 
  5 {
  6        int val;
  7        ListNode *next;
  8        ListNode(int x = 0) : val(x), next(NULL) {}
  9 };
 10  
 11 class Solution 
 12 {
 13 public:
 14     void PrintList(ListNode *pListNode)
 15     {
 16          if (pListNode != NULL) {
 17             cout << pListNode->val << endl;
 18             PrintList(pListNode->next);
 19          } else
 20             cout << "NULL\n" << endl;
 21     }
 22     
 23     int getLengthOfList(const ListNode *pListNode)
 24     {
 25         int length = 0;
 26         
 27         while (pListNode != NULL)
 28         {
 29               pListNode = pListNode->next;
 30               length ++;
 31         }
 32 
 33         return length;              
 34     }
 35     
 36     ListNode* FindFirstCommonNode(ListNode *pHead1, ListNode *pHead2) 
 37     {
 38 //              cout << "Entry ListNode* FindFirstCommonNode(ListNode *pHead1, ListNode *pHead2)" << endl;
 39               
 40               int len1 = getLengthOfList(pHead1);
 41               int len2 = getLengthOfList(pHead2);
 42               
 43               if (len1 > len2)
 44               {
 45                     int gap = len1 - len2;
 46                     for (int i = 0; i < gap; i ++)
 47                         pHead1 = pHead1->next;                       
 48               } 
 49               else 
 50               {
 51                     int gap = len2 - len1;
 52                     for (int i = 0; i < gap; i ++)
 53                        pHead2 = pHead2->next;                                          
 54               }
 55               
 56               while ((pHead1 != NULL) && (pHead2 != NULL) && (pHead1 != pHead2))
 57               {
 58                     pHead1 = pHead1->next;
 59                     pHead2 = pHead2->next;
 60               }
 61 
 62               return pHead1;
 63     }
 64 };
 65 
 66 int main ()
 67 {
 68     Solution testSolution;
 69     int count = 10, k = 0;
 70     
 71     ListNode *pMHead = NULL;
 72     ListNode *pMCur = NULL;
 73     ListNode *pKNode = NULL;
 74     
 75     for (int i = 0; i < count; i ++)
 76     {
 77         ListNode *pTemp = new ListNode(i);
 78         
 79         if (i == 0)
 80             pMHead = pMCur = pTemp;
 81         else {
 82             pMCur->next = pTemp;
 83             pMCur = pMCur->next; // pMCur->next == pTemp
 84         }
 85         
 86         if (i == k)
 87            pKNode = pMCur;
 88     }
 89     
 90     // Print Head       
 91     cout << "pMHead = " << pMHead << endl;
 92     if (pMHead != NULL)
 93        cout << "pMHead->val = " << pMHead->val << endl;
 94     cout << endl;
 95     
 96     // Print Kth Node       
 97     cout << "pKNode = " << pKNode << endl;
 98     if (pKNode != NULL)
 99        cout << "pKNode->val = " << pKNode->val << endl;
100     cout << endl;
101 
102     // Print Linkedlist
103     cout << "PrintList(pMHead)" << endl;
104     testSolution.PrintList(pMHead);        
105     
106     ListNode *pTarget = testSolution.FindFirstCommonNode(pMHead, pKNode);
107     
108     // Print Head       
109     cout << "pMHead = " << pMHead << endl;
110     if (pMHead != NULL)
111        cout << "pMHead->val = " << pMHead->val << endl;
112     cout << endl;
113     
114     // Print Kth Node       
115     cout << "pTarget = " << pTarget << endl;
116     if (pTarget != NULL)
117        cout << "pTarget->val = " << pTarget->val << endl;
118     cout << endl;
119 
120     // Print Linkedlist
121     cout << "PrintList(pMHead)" << endl;
122     testSolution.PrintList(pMHead);        
123 
124     getchar();
125     
126     return 0;
127 }

View Code

posted on 2016-06-11 11:19 浩然119 阅读(528) 评论(0) 编辑收藏举报