leetcode第24题——**Swap Nodes in Pairs

题目

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

题目要求交换链表中的节点对,但不能修改节点值,只是修改节点的前后关系。考察对链表的操作。
用递归实现,详细思路见Java程序的注解。

代码

Python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def swapPairs(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if(head == None or head.next == None):
            return head
            
        nex = head.next
        tmp = nex.next
        
        nex.next = head
        head.next = self.swapPairs(tmp)
        
        return nex

Java

/**
 * 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) {
        //以1->2->3->4举例说明
		if(head == null || head.next == null) return head;
		
		ListNode next = head.next;
		ListNode temp = next.next;//先把3节点开始的链表保存起来,作为下个递归的对象
		
		next.next = head;//2节点的下个节点变成1节点
		head.next = swapPairs(temp);//1节点的下个节点是3节点,递归对3节点开始的链表swapPairs
		
		return next;//2节点开始的链表是要返回的结果
    }
}

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