[算法专题] LinkedList

前段时间在看一本01年出的旧书《effective Tcp/Ip programming》，这个算法专题中断了几天，现在继续写下去。

Introduction

对于单向链表(singly linked list)，每个节点有⼀个next指针指向后一个节点，还有一个成员变量用以储存数值；对于双向链表(Doubly LinkedList)，还有一个prev指针指向前一个节点。与数组类似，搜索链表需要O(n)的时间复杂度，但是链表不能通过常数时间读取第k个数据。链表的优势在于能够以较⾼的效率在任意位置插⼊或删除一个节点。

Dummy Node，Scenario: When the head is not determinated

之前在博客中有几篇文章，已经介绍了Dummy Node的使用，实际上Dummy Node就是所谓的头结点，使用Dummy Node可以简化操作，统一处理head与其他node的操作。

链表操作时利⽤用dummy node是⼀一个⾮非常好⽤用的trick：只要涉及操作head节点，不妨创建dummy node:

ListNode *dummy = new ListNode(0);
dummy->next = head;

废话少说，来看以下几道题：

1. Remove Duplicates from Sorted List I, II
2. Merge Two Sorted Lists
3. Partition List
4. Reverse Linked List I,II

1. Remove Duplicates from Sorted List I

https://leetcode.com/problems/remove-duplicates-from-sorted-list/

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

这题我们的策略遇到重复的node，保留第一个，删除后面的，因此并不会改动head节点，所以下面的代码没有设置dummy node。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if (head == NULL) {
            return head;
        }
        
        ListNode *headBak = head;
        while (head->next) {
            if (head->val == head->next->val) {
                // 保留第一个，删除后面的
                ListNode *tmp = head->next;
                head->next = tmp->next;
                delete tmp;
            } else {
                head = head->next;
            }
        }
        
        return headBak;
    }
};

2. Remove Duplicates from Sorted List II

https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if (head == NULL) {
            return head;
        }
        
        ListNode *dummy = new ListNode(0);
        dummy->next = head;
        
        ListNode *pos = dummy;
        
        // 至少需要head 和 head->next 才可能有重复
        while (pos->next && pos->next->next) {
            if (pos->next->val == pos->next->next->val) {
                int preVal = pos->next->val;
                // 至少保证有两个node，才会有重复
                while (pos->next && pos->next->val == preVal) {
                    ListNode *tmp = pos->next;
                    pos->next = tmp->next;
                    delete tmp;
                }
            } else {
                pos = pos->next;
            }
        }
        
        return dummy->next;
    }
};

3. Merge Two Sorted Lists

https://leetcode.com/problems/merge-two-sorted-lists/

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode *dummy = new ListNode(0);
        ListNode *curr = dummy;
        
        while (l1 && l2) {
            if (l1->val < l2->val) {
                curr->next = l1;
                l1 = l1->next;
            } else {
                curr->next = l2;
                l2 = l2->next;
            }
            
            curr = curr->next;
        }
        
        if (l1) {
            curr->next = l1;
        }
        
        if (l2) {
            curr->next = l2;
        }
        
        return dummy->next;
    }
};

4. Partition List

https://leetcode.com/problems/partition-list/

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        if (head == NULL) {
            return NULL;
        }
        
        ListNode *leftDummy = new ListNode(0);
        ListNode *left = leftDummy;
        
        ListNode *rightDummy = new ListNode(0);
        ListNode *right = rightDummy;
        
        ListNode *curr = head;
        
        while (curr) {
            if (curr->val < x) {
                left->next = curr;
                left = left->next;
                curr = curr->next;
            } else {
                right->next = curr;
                right = right->next;
                curr = curr->next;
            }
        }
        
        left->next = rightDummy->next;
        right->next = NULL;
        
        return leftDummy->next;
    }
};

Basic Skills

5. Reverse Linked List

https://leetcode.com/problems/reverse-linked-list/

Reverse a singly linked list.

Hint:

A linked list can be reversed either iteratively or recursively. Could you implement both?

非递归写法如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (head == NULL) {
            return NULL;
        }
        
        ListNode *pre = NULL;
        ListNode *curr = head;
        ListNode *next = NULL;
        while (curr) {
            next = curr->next;
            
            curr->next = pre;
            pre = curr;
            
            curr = next;
        }
        
        return pre;
    }
};

递归写法如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverseList(ListNode *head) {
        // empty list
        if (head == NULL) return head;
        // Base case
        if (head->next == NULL) return head;
        
        // reverse from the rest after head
        ListNode *newHead = reverseList(head->next);
        // reverse between head and head->next
        head->next->next = head;
        // unlink list from the rest
        head->next = NULL;

        return newHead;
    }
};

6. Reverse Linked List II

https://leetcode.com/problems/reverse-linked-list-ii/

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if (head == NULL) {
            return NULL;
        }
        
        
        ListNode *dummy = new ListNode(0);
        dummy->next = head;
        ListNode *pos = dummy;
        
        for (int i = 0; i <= m - 2; i++) {
            pos = pos->next;
        }
        ListNode *mPreNode = pos;
        ListNode *mCurrNode = pos->next;
        
        
        ListNode *nPreNode = NULL;
        ListNode *nCurrNode = mCurrNode;
        ListNode *nNextNode = NULL;
        for (int i = m; i <= n; i++) {
            nNextNode = nCurrNode->next;
            
            nCurrNode->next = nPreNode;
            nPreNode = nCurrNode;
            
            nCurrNode = nNextNode;
        }
        
        mPreNode->next = nPreNode;
        mCurrNode->next = nCurrNode;
        
        return dummy->next;
    }
};

7. Remove Linked List Elements

https://leetcode.com/problems/remove-linked-list-elements/

Remove all elements from a linked list of integers that have value val.

Example
Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
Return: 1 --> 2 --> 3 --> 4 –> 5

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode *dummy = new ListNode(0);
        dummy->next = head;
        
        ListNode *pre = dummy;
        ListNode *curr = head;
        while (curr) {
            if (curr->val == val) {
                ListNode *waitForDel = curr;
                
                pre->next = curr->next;
                curr = curr->next;
                
                delete waitForDel;
            } else {
                pre = pre->next;
                curr = curr->next;
            }
        }
        
        return dummy->next;
    }
};

8. Remove Nth Node From End of List

https://leetcode.com/problems/remove-nth-node-from-end-of-list/

Given a linked list, remove the n^th node from the end of list and return its head.

For example,
   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode *dummy = new ListNode(0);
        dummy->next = head;
        
        ListNode *slow = dummy;
        ListNode *fast = dummy;
        for (int i = 0; i < n - 1; i++) {
            fast = fast->next;
        }
        
        ListNode *pre = NULL;
        while (fast->next) {
            pre = slow;
            slow = slow->next;
            fast = fast->next;
        }
        
        pre->next = slow->next;
        delete slow;
        
        return dummy->next;
    }
};

9. Remove Duplicates from Sorted List