leetcode -- 24. Swap Nodes in Pairs

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题目描述

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

Example:

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

Your algorithm should use only constant extra space.
You may not modify the values in the list’s nodes, only nodes itself may be changed.
如果可以改变节点的值的话就很简单了。
题目难度：Middle

AC代码

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) { val = x; }* }*/
class Solution {public ListNode swapPairs(ListNode head) {if(head == null || head.next == null) return head;ListNode newHead = new ListNode(0);ListNode curNode = newHead; ListNode temp;while(head != null && head.next != null){temp = head.next.next;curNode.next = head.next;curNode.next.next = head;head = temp;curNode = curNode.next.next;}//if(head != null) curNode.next = head; curNode.next = head; //这一句至关重要。没加的后果：奇数个结点不会出现问题，偶数个结点会使得最后两个结点互指。return newHead.next;}
}

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