[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。