92. 反转链表 II
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92. 反转链表 II
Description
Difficulty: 中等
Related Topics: 链表
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 。
示例 1:
输入:head = [1,2,3,4,5], left = 2, right = 4
输出:[1,4,3,2,5]
示例 2:
输入:head = [5], left = 1, right = 1
输出:[5]
提示:
- 链表中节点数目为
n
1 <= n <= 500
-500 <= Node.val <= 500
1 <= left <= right <= n
进阶: 你可以使用一趟扫描完成反转吗?
Solution
Language: JavaScript
/**
* 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) {
const dummyNode = new ListNode(-1)
dummyNode.next = head // 虚拟节点
let pre = dummyNode
for (let i = 0; i < left - 1; i++) {
// pre 遍历到left前一个节点
pre = pre.next
}
let rightNode = pre
for (let i = 0; i < right - left + 1; i++) {
// rightNode 遍历到right的位置
rightNode = rightNode.next
}
let leftNode = pre.next // 保存leftNode
let curr = rightNode.next // 保存rightNode.next
// 切断left到right的子链
pre.next = null
rightNode.next = null
// 反转left到right的子链
reverseLinkedList(leftNode)
// 反转 连接
pre.next = rightNode
leftNode.next = curr
return dummyNode.next
};
const reverseLinkedList = (head) => {
let pre = null
let cur = head
while (cur) {
const next = cur.next
cur.next = pre
pre = cur
cur = next
}
}