[LeetCode][JavaScript]Reverse Linked List II

Reverse Linked List II

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.

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

---

翻转从m到n的链表。

翻转的操作还是和上一题一样：http://www.cnblogs.com/Liok3187/p/4540490.html

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} m
 * @param {number} n
 * @return {ListNode}
 */
var reverseBetween = function(head, m, n) {
    var res = new ListNode(-1), subHead = res, count = 1;
    res.next = head;
    while(count !== m){
        subHead = head;
        head = head.next;
        count++;
    }
    var subTail = head, tmp;
    while(count !== n + 1){
        tmp = head.next;
        head.next = subHead.next;
        subHead.next = head;
        head = tmp;
        count++;
    }
    subTail.next = head;
    return res.next;
};