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; }
}