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leetcode 【 Swap Nodes in Pairs 】python 实现

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.

 

代码:oj 测试通过 Runtime: 42 ms

             1
            #
             Definition for singly-linked list.
             2
            #
             class ListNode:
             3
            #
                 def __init__(self, x):
             4
            #
                     self.val = x
             5
            #
                     self.next = None
             6
             7
            class
             Solution:

             8
            #
             @param a ListNode
             9
            #
             @return a ListNode
            10
            def
             swapPairs(self, head):

            11
            if head is None or head.next is None: 
12return head
1314         dummyhead = ListNode(0)
15         dummyhead.next = head
1617         pre = dummyhead
18         curr = head
19while curr isnot None and curr.next isnot None:
20             tmp = curr.next
21             curr.next = tmp.next
22             tmp.next = pre.next 
23             pre.next = tmp
24             pre = curr
25             curr = curr.next
26return dummyhead.next

 

思路:

基本的链表操作。

需要注意的是while循环的判断条件:先判断curr不为空,再判断curr.next不为空。

这种and判断条件具有短路功能,如果curr为空就不会进行下一个判断了,因此是安全的

 

原文:http://www.cnblogs.com/xbf9xbf/p/4214359.html


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