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 。
实例1:
输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]
实例2:
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)
* }
*/
function ListNode(val, next) {
this.val = (val===undefined ? 0 : val)
this.next = (next===undefined ? null : next)
}
// 反转链表
const reverseNode = (head) => {
let pre = null;
let q = head;
while(q) {
let temp = q.next;
q.next = pre;
pre = q;
q = temp;
}
}
const reverseBetween = (head, left, right) => {
const resultNode = new ListNode();
resultNode.next = head;
let pre = resultNode;
for (let i = 0; i < left - 1; i++) { // 此循环是为了取到 left 的前一个位置 pre
pre = pre.next;
}
let rightNode = pre;
for (let i = 0; i < right - left + 1; i++) { // 此循环是为了取到 right 位置的指针
rightNode = rightNode.next
}
// 截取 left 至 right 的链表
let newListLeft = pre.next; // 暂存 left 至 right 链表的头指针位置。
let endNode = rightNode.next; // 暂存 right.next 所指的位置
// 截断 left 至 right 指针
pre.next = null;
rightNode.next = null;
reverseNode(newListLeft);
// 反转完后，接回原链表
pre.next = rightNode;
newListLeft.next = endNode; // 反转完后，rightNode 在左端，newListLeft 在右端
return resultNode.next
}function ListNode(val, next) {
this.val = (val===undefined ? 0 : val)
this.next = (next===undefined ? null : next)
}
const reverseBetween = (head, left, right) => {
const resultNode = new ListNode(-1);
resultNode.next = head;
let pre = resultNode
for(let i = 0; i < left - 1; i++) {
pre = pre.next;
}
let cur = pre.next;
for(let i = 0; i < right - left; i++) {
let q = cur.next;
cur.next = q.next;
q.next = pre.next;
pre.next = q;
}
return resultNode.next
}