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LeetCode-Swap Nodes in Pairs

时间:2016-05-03 01:56:47      阅读:126      评论:0      收藏:0      [点我收藏+]

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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.

这道题考验链表的遍历, 切记要加上一个dummy head, 并且区分index的单偶数,不难

/**
 * 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 dummy=new ListNode(0);
        dummy.next=head;
        ListNode pd=dummy;
        int i=1;
        while(pd!=null){
            if(i%2 != 0){
                if(pd.next != null && pd.next.next != null){
                    ListNode temp1 = pd.next.next.next;
                    ListNode temp2 =pd.next;
                    pd.next=pd.next.next;
                    pd.next.next=temp2;
                    pd.next.next.next=temp1;
                }
            }
            pd=pd.next;
            i++;
        }
        return dummy.next;
    }
}

 

LeetCode-Swap Nodes in Pairs

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原文地址:http://www.cnblogs.com/incrediblechangshuo/p/5453656.html

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