00链表中等 牛客NC.2重排链表 leetcode143. 重排链表_将给定的单链表l:l0→l1→l2→…→ln-2→ln-1→ln

143. 重排链表

题目描述

给定一个单链表 L：L0→L1→…→Ln-1→Ln ，
将其重新排列后变为： L0→Ln→L1→Ln-1→L2→Ln-2→…
你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例 1:
给定链表 1->2->3->4, 重新排列为 1->4->2->3.
示例 2:
给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.

双端队列解法

import java.util.Deque;
import java.util.LinkedList;
class Solution {
    public void reorderList(ListNode head) {
        if(head == null || head.next == null){
            return ;
        }
        Deque<ListNode> dequeue = new LinkedList<>();
        ListNode cur = head;
        while(cur != null){
            dequeue.offer(cur);
            cur = cur.next;
        }

        while(!dequeue.isEmpty()){
            head.next = dequeue.pollFirst();
            head = head.next;
            if(!dequeue.isEmpty()){
                head.next = dequeue.pollLast();
                head = head.next;
            }
        }
        head.next = null;
    }
}
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结点指针移动

代码

import java.util.Deque;
import java.util.LinkedList;
class Solution {
    public void reorderList(ListNode head) {
        if(head == null || head.next == null){
            return ;
        }
        //找到整个链表的中心点
        ListNode fast = head;
        ListNode low = head;
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            low = low.next;
        }
        
        //反转后半段链表
        ListNode pre = null;
        ListNode cur = low;
        ListNode temp = cur.next;
        cur.next = pre;
        cur = temp;
        while(cur != null){
            temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
        
        //前后两段链表混合
        while(pre != null){
            ListNode temp1 = head.next;
            ListNode temp2 = pre.next;
            head.next = pre;
            pre.next = temp1;
            head = temp1;
            pre = temp2;
        }
    }
}
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出错代码

import java.util.Deque;
import java.util.LinkedList;
class Solution {
    public void reorderList(ListNode head) {
        if(head == null || head.next == null){
            return ;
        }
        ListNode fast = head;
        ListNode low = head;
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            low = low.next;
        }
        if(fast != null){
            low = low.next;
        }
        
        //反转链表。这里错了，造成pre最后一个结点无限循环。 但我不明白，为什么有错？
        ListNode pre = null;
        ListNode cur = low;
        while(cur != null){
            ListNode temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
        
        while(pre != null){
            ListNode temp1 = head.next;
            ListNode temp2 = pre.next;
            head.next = pre;
            pre.next = temp1;
            head = temp1;
            pre = temp2;
        }
    }
}
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