LeetCode：92. Reverse Linked List II

一、问题介绍

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

二、思路解析

对于链表的问题，需要建一个preHead，连上原链表的头结点，这样的话就算头结点变动了，我们还可以通过preHead->next来获得新链表的头结点。

 

三、代码实现

/**
 * 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) {
       ListNode *dummy = new ListNode(0), *pre = dummy;
        dummy->next = head;
        for (int i = 1; i <= m - 1; ++i) pre = pre->next;
        ListNode *cur = pre->next;
        for (int i = m; i <= n - 1; ++i) {
            ListNode *t = cur->next;
            cur->next = t->next;
            t->next = pre->next;
            pre->next = t;
        }
        return dummy->next;
    }
};

 

来源：https://blog.csdn.net/ixiaowei1993/article/details/100935222