码迷,mamicode.com
首页 > 其他好文 > 详细

24. Swap Nodes in Pairs

时间:2015-05-14 08:36:05      阅读:88      评论:0      收藏:0      [点我收藏+]

标签:

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.

实现题,注意边界情况即可

/**
* 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 newHead = new ListNode(0);
    newHead.next = head;
    head = newHead;
    while (head!= null && head.next != null && head.next.next != null) {
      ListNode next = head.next;
      head.next = head.next.next;
      next.next = head.next.next;
      head.next.next = next;
      head = head.next.next;
    }
    return newHead.next;
  }
}

24. Swap Nodes in Pairs

标签:

原文地址:http://www.cnblogs.com/shini/p/4502322.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!