【LeetCode】24. Swap Nodes in Pairs 两两交换链表中的节点

【LeetCode】Swap Nodes in Pairs 解题报告

标签（空格分隔）： LeetCode

题目地址：​​https://leetcode.com/problems/swap-nodes-in-pairs/description/​​

题目描述：

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.

解题方法

交换相邻的两个链表元素。

pre -> a -> b -> b.next 转换成 pre -> b -> a -> b.next，需要修改3个指针。

python 代码如下：

# 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
        """
        pre, pre.next = self, head
        while pre.next and pre.next.next:
            a = pre.next
            b = a.next
            pre.next, a.next, b.next = b, b.next, a
            pre = a
        return self.next

java 代码如下：

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

日期

2017 年 8 月 29 日