leetCode 24. Swap Nodes in Pairs (双数交换节点) 解题思路和方法

标签：leetcode   链表   

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

思路：题目比较简单，会链表反转的都可以做，思路也差不多，不多说，上代码。

/**
 * 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) {
	        
	        ListNode firstHead = new ListNode(0);
	        firstHead.next = head;
	        ListNode pre = firstHead;//定义一个头结点，这样所有的操作都相同
	        ListNode next = null;
	        while(head != null && head.next != null){
	        	//A-B-C-D交换BC,pre=A;B=head;C=next
	            next = head.next;//保存交换的变量C
	            
	            head.next = next.next;//将B指向B的指针指向D
	            pre.next = next;//将A指向B的指针指向C
	            next.next = head;//将C指向D的指针指向B,完成交换，顺序变为A-C-B-D
	            
	            //为下一循环准备变量
	            pre = head;//将pre变为B
	            head = head.next;//将head指向D
	            
	        }
	        return firstHead.next;
	    }
}

leetCode 24. Swap Nodes in Pairs (双数交换节点) 解题思路和方法

标签：leetcode   链表   

原文地址：http://blog.csdn.net/xygy8860/article/details/46788099