LeetCode之旅（21）-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.

思路：

• 首先是判断head和head.next为空的情况，直接返回head
• 然后设置first和second两个变量，用来记录一前一后两个元素
• 设置pre来记录上一次最后的元素。pre.next = second；
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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||head.next == null){            return head;        }        ListNode first = head;        ListNode second = head.next;        ListNode pre = null;        ListNode swap = null;        if(second != null){            head = second;        }        while(first != null&&second != null){            swap = second.next;            second.next = first;            first.next = swap;            if(pre != null){                pre.next = second;            }            pre = first;            first = swap;            if(swap != null){                second = swap.next;            }        }        return head;    }}