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

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param a ListNode
    # @return a ListNode
    def swapPairs(self, head):
        if head is None or head.next is None: 
            return head
        
        dummyhead = ListNode(0)
        dummyhead.next = head
        
        pre = dummyhead
        curr = head
        while curr is not None and curr.next is not None:
            tmp = curr.next
            curr.next = tmp.next
            tmp.next = pre.next 
            pre.next = tmp
            pre = curr
            curr = curr.next
        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