[LeetCode] 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.

/**
 * 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 prev = null;
    var now = head;
    var next = head.next;
    var iprev = null;
    var inext = null;
    var ihead = null;
    var itail = null;
    var s = 1;

    var newhead = false;

    while (now) {
        if (s >= m && s <= n) {
            if (s == m) {
                itail = now;
                iprev = prev;
            }
            if (s == n) {
                ihead = now;
                inext = next;
            }
            now.next = prev;
            prev = now;
            now = next;
            if (now) {
                next = now.next;
            }

            if (s == n) {
                if (!iprev) {
                    head = ihead;
                } else {
                    iprev.next = ihead;
                }
                itail.next = inext;
                break;
            }
        } else {
            prev = now;
            now = next;
            if (now) {
                next = now.next;
            }
        }
        s++;
    }
    return head;
};

代码比较丑，用了大量的零时变量来存东西，思路跟翻转整个链表那题如出一辙。简述一下思路就是：用三指针prev， now， next 遍历链表，当到达指定范围后开始翻转操作，并顺便记录翻转那部分的头和尾（ihead, itail）还有相应接头部分的指针（iprev, inext）以便最后翻转完了以后把两部分接起来。特别注意的是链表的头可能会被改变。