题目描述

• 一看题目反序：用栈
• 更新：O(1) 空间复杂度
  

思路 & 代码

• 先快慢指针，找到需要入栈的起点，然后逐个入栈
• 然后逐个出栈，进行重排即可
• 注意1：记得对重排后的链表结尾进行 Last.next = null 处理，防止成环
• 注意2：奇偶情况要进行处理，可见代码注释
• 时间复杂度 O(n) ，空间复杂度 O(n)

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public void reorderList(ListNode head) {// 快慢指针ListNode now = head;// 走到需要进行入栈的位置now = head;ListNode pre = head;while(now.next != null && now.next.next != null){pre = pre.next;now = now.next.next;}// 偶数处理if(now.next != null){pre = pre.next;}now = pre.next;// 末尾处理，防止死循环pre.next = null;// 入栈Stack<ListNode> myStack = new Stack<>();while(now != null){myStack.push(now);now = now.next;}// 重排now = head;while(!myStack.isEmpty()){ListNode temp = myStack.pop();ListNode nextLoop = now.next;temp.next = now.next;now.next = temp;now = nextLoop;}}
}

更新版：快慢指针 + 翻转链表

class Solution {public void reorderList(ListNode head) {// 0. 特殊情况if(head == null || head.next == null) {return;}// 1. 快慢指针，划分出两段（注意第一段结尾的 next 置空，用 slowPreListNode slow = head, fast = head;ListNode slowPre = null;while(fast != null && fast.next != null) {slowPre = slow;slow = slow.next;fast = fast.next.next;}// 奇数，slow 为中间节点（奇偶情况处理）if(fast != null) {slowPre = slow;slow = slow.next;}// 2. 第二段的翻转slowPre.next = null;ListNode pre = null;while(slow != null) {ListNode nextNode = slow.next;slow.next = pre;pre = slow;slow = nextNode;}// 3. 合并两段链表fast = head;while(pre != null) {ListNode temp1 = fast.next;ListNode temp2 = pre.next;fast.next = pre;pre.next = temp1;fast = temp1;pre = temp2;}}
}