LeetCode--Reverse Linked List II

题目链接：

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

思路：

由于题目给的条件比较宽松，需要考虑的几种情况就是：m=n；m=1；n=lenght;其他m<n

先找到第m-1个节点，以及第m个节点，从m+1遍历到n，依次插入到m-1个节点的后面

8ms

/**
 * 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 || head->next == NULL)
            return head;
            
        ListNode * pre = new ListNode(-1);  
        pre->next = head;  
          
        //pre指向head的前一个  
        ListNode *current = pre;  
          
        //把current指向第m个元素的前一个  
        for(int i = 1 ;i < m ;i++){  
            current = current->next;  
        }  
        //revHead指向第M个节点，revCur用来遍历M+1到n的节点，插入到第m-1
        //个节点的后面
        ListNode *revHead = current->next;
        ListNode *revCur = revHead->next;
        for(int j = m+1 ; j <= n ; ++j)
        {
            ListNode *tmp = revCur->next;
            revCur->next = current->next;
            current->next = revCur;
            revCur = tmp;
        }
        revHead->next = revCur;
        return pre->next;
    }
};