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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 ListNode10     def swapPairs(self, head):11         if head is None or head.next is None: 12             return head13         14         dummyhead = ListNode(0)15         dummyhead.next = head16         17         pre = dummyhead18         curr = head19         while curr is not None and curr.next is not None:20             tmp = curr.next21             curr.next = tmp.next22             tmp.next = pre.next 23             pre.next = tmp24             pre = curr25             curr = curr.next26         return 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

 

leetcode 【 Swap Nodes in Pairs 】python 实现