55 lines
1.5 KiB
Markdown
55 lines
1.5 KiB
Markdown
方法一：快慢指针 + 反转链表 + 合并链表
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> 作者：lcbin
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链接：https://leetcode.cn/problems/reorder-list/solution/python3javacgo-yi-ti-yi-jie-kuai-man-zhi-t9u2/
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来源：力扣（LeetCode）
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著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
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我们先用快慢指针找到链表的中点，然后将链表的后半部分反转，最后将左右两个链表合并。
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```js
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/**
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* Definition for singly-linked list.
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* function ListNode(val, next) {
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* this.val = (val===undefined ? 0 : val)
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* this.next = (next===undefined ? null : next)
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* }
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*/
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/**
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* @param {ListNode} head
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* @return {void} Do not return anything, modify head in-place instead.
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*/
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var reorderList = function (head) {
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// 快慢指针找到链表中点
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let slow = head;
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let fast = head;
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while (fast.next && fast.next.next) {
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slow = slow.next;
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fast = fast.next.next;
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}
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// cur 指向右半部分链表
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let cur = slow.next;
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slow.next = null;
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// 反转右半部分链表
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let pre = null;
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while (cur) {
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const t = cur.next;
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cur.next = pre;
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pre = cur;
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cur = t;
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}
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cur = head;
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// 此时 cur, pre 分别指向链表左右两半的第一个节点
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// 合并
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while (pre) {
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const t = pre.next;
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pre.next = cur.next;
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cur.next = pre;
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cur = pre.next;
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pre = t;
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}
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};
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```
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时间复杂度O(n)，其中 n 是链表的长度。空间复杂度O(1) |