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；
-

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;
    }
}