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.

思路:

递归。要搞清楚怎么交换,保证不被更新掉,画张图。1.next=2.next; 2.next=1.next 1.next=swap(1.next)

 

/**
 * 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) {
    if(head==null||head.next==null){
        return head;
    }
    ListNode res=head.next;
    head.next=res.next;
    res.next=head;
    head.next=swapPairs(head.next);
    return res;
    }
}

 

posted on 2016-10-08 14:39  Machelsky  阅读(117)  评论(0)    收藏  举报