leetcode Swap Nodes in Pairs 2.10 难度系数2

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本文介绍了一种链表操作算法，实现每两个相邻节点的交换，并使用常量空间复杂度，不修改节点值只调整指针。

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Question:

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.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
   public static ListNode swapPairs(ListNode head) {
		if(head==null) return null;
		if(head.next==null) return head;
		ListNode first = head;
		ListNode second = first.next;
		ListNode third = second.next;
		
		head = head.next;
		while(second!=null){
			second.next = first;
			if(third!=null&&third.next!=null){
				first.next = third.next;
			}else {
				first.next = third;
			}
			first = third;
			if (third!=null) {
				second = third.next;
			}else {
				second = null;
			}
			if(third!=null&&third.next!=null){
				third = third.next.next;
			}else{
				third = null;
			}
		}
		return head;
	}
}