一道关于链表反转的面试题

题目描述

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

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

/**
 * 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) {
        
        ////
    }
};

实现代码

  ListNode* reverseBetween(ListNode* head, int m, int n) {
        
        ListNode *tmpHead = new ListNode();
        tmpHead->next = head;
        ListNode *pre = tmpHead;

        for(int i = 1; i < m; i++){
            pre = pre->next;
        }

        head = pre->next;
        for(int i = m; i < n; i++){
            ListNode *post = head->next;
            head->next = post->next;
            post->next = pre->next;
            pre->next = post;
        }
        return tmpHead->next;
  }

其中，pre 这个指针一直不动，head 和 post 逐步后移，直至 i < n 为止。完成并返回 tmpHead->next 。