[LeetCode] 24. 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.

 

解法:

  循环的方式:

/**
 * 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;
        
        head = dummy;
        while (head.next != null && head.next.next != null) {
            ListNode left = head.next;
            ListNode right = left.next;
            
            // head -> left -> right -> ....
            // to: head -> right -> left -> ....
            left.next = right.next;
            right.next = left;
            head.next = right;
            head = left;
        }
        
        return dummy.next;
    }
}

 

  递归的方式:

/**
 * 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 temp = head.next;
        head.next = swapPairs(temp.next);
        temp.next = head;
        return temp;
    }
}

 

posted @ 2017-02-20 14:27  Strugglion  阅读(121)  评论(0编辑  收藏  举报