92. Reverse Linked List II

1. 问题描述
   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.

2. 解决思路
   首先找到需要reverse的区间段，然后依次将前面的插入到后面即可。需要主要的是，这里可以在head前面弄多一个节点，可以巧妙处理m=1的情况。

3. 代码

/** * 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 (m == n)            return head;        ListNode *pre = new ListNode(0);        pre->next = head;        ListNode *front_cur, *front_pre, *back_cur,*back_pre,*tmp,*cur = head,*begin = pre;        int count = 0;        while(cur != NULL) {            ++count;            if (count == m) {                front_cur = cur;                front_pre = pre;            }            if (count == n) {                back_cur = cur;                back_pre = pre;                break;            }            pre = cur;            cur = cur->next;        }        count = 0;        while(count++ < n-m) {            front_pre->next = front_cur->next;            front_cur->next = back_cur->next;            back_cur->next = front_cur;            front_cur = front_pre->next;        }        return begin->next;    }};