leetCode:Swap Nodes in Pairs

Swap Nodes in Pairs

题目

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

交换相邻节点，但不允许改变节点的值

思路

通过改变每个节点的指针指向来交换相邻节点。

分以下情况

1. 如果相邻节点的后面没有节点
2. 相邻节点后面仅剩一个节点
3. 相邻节点后面还有大于等于2个节点

AC代码

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode swapPairs(ListNode head) {        if(head == null)             return null;        if(head.next == null){            return head;        }        ListNode newHead = head.next;        while(head.next != null){            ListNode nextNode = head.next;            //如果相邻节点的后面没有节点            if(nextNode.next == null){                head.next = null;            }else{                //相邻节点后面仅剩一个节点                if(nextNode.next.next == null){                    head.next = nextNode.next;                }                //相邻节点后面还有大于等于2个节点                else{                    head.next = nextNode.next.next;                }            }            ListNode temp = nextNode.next;            nextNode.next = head;            //移动head，指向相邻节点的下一个节点            if(temp != null){                head = temp;            }else{                break;            }        }        return newHead;    }}