LeetCode--reverse-linked-list-ii

题目描述

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

For example:

Given1->2->3->4->5->NULL, m = 2 and n = 4,

return1->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) {}
* };
*/
//分为三部分，m到n之间的数据反转后，与前后的链表重新连接
class Solution {
public:
     
    ListNode *reverseBetween(ListNode *head, int m, int n)
    {    
        ListNode *fake = new ListNode(0); //构造一个假头结点，防止m=1时，翻转链表后找不到整个链表的头结点
        fake->next = head;
        ListNode *p = fake;
         
        for(int i=0;i<m-1;i++)
            p = p->next;
        ListNode *firstTail = p,*secondTail = p->next;
//firstTail指向m-1，secondTail指向m
         
        ListNode *pre = nullptr,*node = secondTail;//头插法翻转链表
        for(int i=0;i<n-m+1;i++)
         {
            ListNode *temp = node->next;
            node->next = pre;
            pre = node;
            node = temp;
         }
         firstTail->next = pre;
         secondTail->next = node; //三段链表的连接
         
        return fake->next;
    }
};