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leetcode Linked List Swap Nodes in Pairs python 实现

题目:

Swap Nodes in Pairs

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164ms通过

 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         12         if head is None or head.next is None:13             return head14         15         dummyhead = ListNode(0)16         dummyhead.next = head17         p = dummyhead18         19         while p.next is not None and p.next.next is not None:20             tmp = p.next.next.next21             p.next.next.next = p.next22             p.next = p.next.next23             p.next.next.next = tmp24             p = p.next.next25             26             27         return dummyhead.next

 

思路

1. 建立一个虚表头hummyhead 这样方便操作一些

2. p.next始终指向要交换的下一个元素的位置

3. 先保存p.next.next.next,再移动p,p.next,p.next.next,p.next.next.next:先动p.next.next.next再动其他的。

小白我一开始先动的是p,p.next结果后面的p.next.next就丢了,其他小白别陷入这个误区,高手请略过。

Tips: 动了哪个指针,就把哪个指针上面打个×;添加了哪个指针,就在两个点之间加一根线;画画图就出来了,别光看着不动笔。

 

leetcode Linked List Swap Nodes in Pairs python 实现