24. Swap Nodes in Pairs [easy] (Python)

题目链接

https://leetcode.com/problems/swap-nodes-in-pairs/

题目原文

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.

题目翻译

给定一个单链表，交换相邻的节点，并返回链表头。比如，给定1->2->3->4，你应该返回 2->1->4->3。
注意：你的算法应该是O(1)的空间复杂度，不能修改链表节点的值（val），只能修改节点。

思路方法

思路一

既然每次交换相邻节点，那么就以两个节点为一个单位做处理，循环每次交换两个相邻节点即可。

代码

# 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 not head or not head.next:
            return head
        pre = new_head = ListNode(0)
        while head and head.next:
            tmp = head.next
            head.next = tmp.next
            tmp.next = head
            pre.next = tmp
            pre = head
            head = head.next
        return new_head.next

说明
上面的思路，可以换个角度考虑：现在是用现在的节点构造一个新的链表，每次向新链表添加两个节点，这两个节点是原链表的相邻节点的逆序。

思路二

代码

# 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 not head or not head.next:
            return head
        new_head = head.next
        head.next = self.swapPairs(head.next.next)
        new_head.next = head
        return new_head

PS: 新手刷LeetCode，新手写博客，写错了或者写的不清楚还请帮忙指出，谢谢！
转载请注明：http://blog.csdn.net/coder_orz/article/details/51532184