[leetcode]Swap Nodes in Pairs @ Python

 

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.

 

题意：将链表中的节点两两交换。Given 1->2->3->4, you should return the list as 2->1->4->3.

 

解题思路：这题主要涉及到链表的节本操作。加一个头结点，操作起来会很方便。另外配了一个示意图　［本图是我asrman原创］

 

 

 

代码：

 

复制代码

# 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):

        pre = ListNode(0)

        pre.next = head

        curr = head

        head = pre

        while curr and curr.next:      # curr =1, curr.next =2

            pre.next = curr.next       # 0 --> 2

            curr.next = pre.next.next  # 1 --> 3  # curr.next.next

            pre.next.next = curr       # 3 --> 1

            pre = curr                 # pre = 1

            curr = curr.next           # curr= 3

        return head.next