给你单链表的头指针 head 和两个整数 left 和 right ,其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点,返回 反转后的链表 。
输入: head = [1,2,3,4,5], left = 2, right = 4
输出: [1,4,3,2,5]
输入: head = [5], left = 1, right = 1
输出: [5]
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @param {number} left
* @param {number} right
* @return {ListNode}
*/
var reverseBetween = function(head, left, right) {
if (!head) return null;
let ret =new ListNode(-1,head),pre = ret,cnt = right-left+1
while(--left){
pre = pre.next
}
pre.next = reverse(pre.next,cnt)
return ret.next;
};
var reverse = function (head, n) {
let pre = null,cur = head;
while(n--) {
[cur.next, pre, cur] = [pre,cur,cur.next]
}
head.next = cur
return pre
}