将大问题分解,先将第一个节点拿出来,将其它的节点看成一个整体。

 1 #include <iostream>
 2 #include <cstring>
 3 #include "DTString.h"
 4 
 5 using namespace std;
 6 using namespace DTLib;
 7 
 8 struct Node
 9 {
10     int value;
11     Node* next;
12 };
13 
14 Node* create_list(int v, int len)  // v:数据元素从哪一个之开始。 len:长度
15 {
16     Node* ret = NULL;
17     Node* slider = NULL;
18 
19     for(int i=0; i<len; i++)
20     {
21         Node* n = new Node();
22 
23         n->value = v++;
24         n->next = NULL;
25 
26         if( slider == NULL )
27         {
28             slider = n;
29             ret = n;
30         }
31         else
32         {
33             slider->next = n;
34             slider = n;
35         }
36     }
37 
38     return ret;
39 }
40 
41 void destroy_list(Node* list)
42 {
43     while( list )
44     {
45         Node* del = list;
46 
47         list = list->next;
48 
49         delete del;
50     }
51 }
52 
53 void print_list(Node* list)
54 {
55     while( list )
56     {
57         cout << list->value << "->";
58 
59         list = list->next;
60     }
61 
62     cout << "NULL" << endl;
63 }
64 
65 Node* reverse(Node* list)
66 {
67     if( (list == NULL) || (list->next == NULL) )
68     {
69         return list;
70     }
71     else
72     {
73         Node* guard = list->next;
74         Node* ret = reverse(list->next);
75 
76         guard->next = list;
77 
78         list->next = NULL;
79 
80         return ret;
81     }
82 }
83 
84 int main()
85 {
86     Node* list = create_list(1, 5);
87 
88     print_list(list);
89 
90     list = reverse(list);
91 
92     print_list(list);
93 
94     destroy_list(list);
95 
96     return 0;
97 }

 

 

实验2:

 

 

 

  1 #include <iostream>
  2 #include <cstring>
  3 #include "DTString.h"
  4 
  5 using namespace std;
  6 using namespace DTLib;
  7 
  8 struct Node
  9 {
 10     int value;
 11     Node* next;
 12 };
 13 
 14 Node* create_list(int v, int len)  // v:数据元素从哪一个之开始。 len:长度
 15 {
 16     Node* ret = NULL;
 17     Node* slider = NULL;
 18 
 19     for(int i=0; i<len; i++)
 20     {
 21         Node* n = new Node();
 22 
 23         n->value = v++;
 24         n->next = NULL;
 25 
 26         if( slider == NULL )
 27         {
 28             slider = n;
 29             ret = n;
 30         }
 31         else
 32         {
 33             slider->next = n;
 34             slider = n;
 35         }
 36     }
 37 
 38     return ret;
 39 }
 40 
 41 void destroy_list(Node* list)
 42 {
 43     while( list )
 44     {
 45         Node* del = list;
 46 
 47         list = list->next;
 48 
 49         delete del;
 50     }
 51 }
 52 
 53 void print_list(Node* list)
 54 {
 55     while( list )
 56     {
 57         cout << list->value << "->";
 58 
 59         list = list->next;
 60     }
 61 
 62     cout << "NULL" << endl;
 63 }
 64 
 65 Node* reverse(Node* list)
 66 {
 67     if( (list == NULL) || (list->next == NULL) )
 68     {
 69         return list;
 70     }
 71     else
 72     {
 73         Node* guard = list->next;
 74         Node* ret = reverse(list->next);
 75 
 76         guard->next = list;
 77 
 78         list->next = NULL;
 79 
 80         return ret;
 81     }
 82 }
 83 
 84 Node* merge(Node* list1, Node* list2)
 85 {
 86     if( list1 == NULL )
 87     {
 88         return list2;
 89     }
 90     else if( list2 == NULL )
 91     {
 92         return list1;
 93     }
 94     else if( list1->value < list2->value )
 95     {
 96         /*
 97         Node* list1_ = list1->next;
 98         Node* list = merge(list1_, list2);
 99 
100         list1->next = list;
101         return list1;
102         */
103         return (list1->next = merge(list1->next, list2), list1); //逗号表达式
104     }
105     else
106     {
107         /*
108         Node* list2_ = list2->next;
109         Node* list = merge(list1, list2_);
110 
111         list2->next = list;
112 
113         return list2;
114         */
115 
116         return (list2->next = merge(list2->next, list1), list2); //逗号表达式
117     }
118 }
119 
120 int main()
121 {
122     Node* list1 = create_list(1, 5);
123 
124     Node* list2 = create_list(2, 6);
125 
126     print_list(list1);
127     print_list(list2);
128 
129     Node* list = merge(list1, list2);
130 
131     print_list(list);
132 
133     destroy_list(list);
134 
135     return 0;
136 }

 

 

 

 

 1 #include <iostream>
 2 #include <cstring>
 3 #include "DTString.h"
 4 
 5 using namespace std;
 6 using namespace DTLib;
 7 
 8 
 9 void HanoiTower(int n, char a, char b, char c) // a ==> src  b ==> middle  c ==> dest
10 {
11     if( n == 1 )
12     {
13         cout << a << "-->" << c << endl;
14     }
15     else
16     {
17         HanoiTower(n-1, a, c, b);
18         HanoiTower(1, a, b, c);
19         HanoiTower(n-1, b, a, c);
20     }
21 }
22 
23 int main()
24 {
25     HanoiTower(3, 'a', 'b', 'c');
26 
27     return 0;
28 }

 

 

 

 

 

 e始终指向字符串的开头,用来打印,s用来控制交换。

一开始,a和a交换,全排列b、c。

然后,a和b交换,全排列a、c。

然后交换a和c,全排列b、a。

如果两个字符是一样的,就没有必要交换,否则全排列有重复的现象。

 1 #include <iostream>
 2 #include <cstring>
 3 #include "DTString.h"
 4 
 5 using namespace std;
 6 using namespace DTLib;
 7 
 8 
 9 void HanoiTower(int n, char a, char b, char c) // a ==> src  b ==> middle  c ==> dest
10 {
11     if( n == 1 )
12     {
13         cout << a << "-->" << c << endl;
14     }
15     else
16     {
17         HanoiTower(n-1, a, c, b);
18         HanoiTower(1, a, b, c);
19         HanoiTower(n-1, b, a, c);
20     }
21 }
22 
23 void permutation(char* s, char* e)  // e始终指向字符串开头,用于打印
24 {
25     if( *s == '\0' )
26     {
27         cout << e << endl;
28     }
29     else
30     {
31         int len = strlen(s);
32         for(int i=0; i<len; i++) //第一个字符一次和后面的元素交换
33         {
34             if( (i == 0) || (s[0] != s[i]) )
35             {
36                 swap(s[0], s[i]);   // 交换
37                 permutation(s+1, e); // 交换之后将子串全排列
38                 swap(s[0], s[i]); // 再交换回来
39             }
40         }
41     }
42 }
43 
44 int main()
45 {
46     char s[] = "abc";
47     char s1[] = "aac";
48 
49     permutation(s, s);
50     cout << "----------" << endl;
51     permutation(s1, s1);
52 
53     return 0;
54 }

 

posted on 2018-09-19 22:42  周伯通789  阅读(238)  评论(0编辑  收藏  举报