LeetCode 92. Reverse Linked List II

Description:

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.

Submission Details
44 / 44 test cases passed.
Status: Accepted
Runtime: 3 ms

/** * 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* h = head;        ListNode* t = head;        ListNode* left;        ListNode* right;        for (int i = 1; i < m; i++) {            left = h; //翻转序列的前一个            h = h->next; //定位到翻转的第一个        }        for (int i = 1; i < n; i++) {            t = t->next; //定位到翻转的最后一个        }        right = t->next; //翻转序列的下一个        ListNode* pre = right;        ListNode* cur = h;        while (cur != right) {            ListNode* tmp = cur->next;            cur->next = pre;            pre = cur;            cur = tmp;        }        h->next = right;        if (m == 1)             return pre;        left->next = pre;        return head;    }};