leetcode 24. Swap Nodes in Pairs-蒲公英云

leetcode 24. Swap Nodes in Pairs

Dear 丶 2023-07-17 15:47 23阅读 0赞

一、问题描述

二、代码实现

1、独立处理第一组节点

2、引入伪头节点使得处理形式统一起来

3、递归版本

https://leetcode.com/problems/swap-nodes-in-pairs/

将单链表每两个节点互换位置。注意不能只改变节点的值，节点必须实实在在地进行交换。

一、问题描述

测试用例：

Example:
Given 1->2->3->4, you should return the list as 2->1->4->3.

需要区分节点个数为偶数和奇数的情况，节点个数为奇数时最后一个节点位置不变。

二、代码实现

1、独立处理第一组节点

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode swapPairs1(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        // the size of list is >= 2
        // swap the first 2 nodes
        ListNode temp = head;
        head = head.next;
        temp.next = head.next;
        head.next = temp;
        ListNode pre = temp;
        while (pre.next != null && pre.next.next != null) {
            ListNode first = pre.next;
            ListNode second = pre.next.next;
            pre.next = second;
            first.next = second.next;
            second.next = first;
            //pre -> second -> first
            pre = pre.next.next;//pre = first;
        }
        return head;
    }
}

2、引入伪头节点使得处理形式统一起来

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode swapPairs2(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode tempHead = new ListNode(-1);
        tempHead.next = head;
        ListNode pre = tempHead, cur = head;
        while (cur != null && cur.next != null) {
            ListNode next = cur.next.next;
            pre.next = cur.next;
            cur.next = next;
            pre.next.next = cur;
            pre = cur;
            cur = next; //cur = cur.next;
        }
        //cur == null（没有节点了） or cur != null && cur.next == null（只有一个节点）
        return tempHead.next;
    }
}

3、递归版本

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode swapPairs3(ListNode head) {
        //1、出口
        if (head == null || head.next == null) {
            return head;
        }
        ListNode newHead = head.next;
        head.next = swapPairs(head.next.next);
        newHead.next = head;
        return newHead;
    }
}