Swap Nodes in Pairs @LeetCode

判断比较花时间，需要分奇数个节点和偶数个节点的情况

package Level2;

import Utility.ListNode;

/**
 * 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.
 *
 */
public class S24 {

	public static void main(String[] args) {
		ListNode n1 = new ListNode(1);
		ListNode n2 = new ListNode(2);
		n1.next = n2;
		ListNode n3 = new ListNode(3);
		n2.next = n3;
		n1.print();
		ListNode head = swapPairs(n1);
		head.print();
	}
	
	public static ListNode swapPairs(ListNode head) {
        if(head == null){
        	return null;
        }
        // 当只有一个元素的情况
        if(head.next == null){
        	return head;
        }
        ListNode i = head;	// i指向第1个
        ListNode j = i.next;	// j指向第2个
        ListNode k = j.next;	// k指向第3个
        
        head = head.next;	
        while(j != null){
        	j.next = i;
        	if(k!=null && k.next!=null){	// 当有偶数个节点 
        		i.next = k.next;
        	}else{		// 当有奇数个节点
        		i.next = k;
        	}
        	
        	// 更新i，j，k的值，前进两格
        	i = k;
        	if(k != null){
        		j = k.next;
        	}else{
        		j = null;
        	}
        	if(k!=null && k.next!=null){
        		k = k.next.next;
        	}else{
        		k = null;
        	}
        }
        return head;
	}
}