LeetCode----- 24.Swap Nodes in Pairs

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.

给定一个链表，交换每两个相邻的节点并返回其头。例如，给定1-> 2-> 3-> 4，您应该将列表作为2-> 1-> 4-> 3返回。您的算法应该仅使用恒定空间。 您不能修改列表中的值，只能改变节点本身。

代码如下：

public class SwapNodesinPairs {
public static ListNode swapPairs(ListNode head) {
if(head == null || head.next == null) {
return head;
}
ListNode node = head;
ListNode temp = null;
int length = 0;
while(node != null) {
length++;
if(length%2 == 0) {
int flag = temp.val;
temp.val = node.val;
node.val = flag;
}else {
temp = node;
}
node = node.next;
}
return head;
}

public static void main(String[] args) {
ListNode l10 = new ListNode(1);
ListNode l11 = new ListNode(2);
ListNode l12 = new ListNode(3);
ListNode l13 = new ListNode(4);
ListNode l14 = new ListNode(5);
ListNode l15 = new ListNode(6);
l10.next = l11;
l11.next = l12;
l12.next = l13;
l13.next = l14;
l14.next = l15;
l15.next = null;

ListNode node =swapPairs(l10);
while(node != null) {
if(node.next == null) {
System.out.println(node.val);
}else{
System.out.print(node.val +"->");
}
node = node.next;
}
}
}

class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}