【python3】leetcode 24. Swap Nodes in Pairs (Medium)

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24. Swap Nodes in Pairs (Medium)

Given a linked list, swap every two adjacent nodes and return its head.

Example:

Given 1->2->3->4, you should return the list as 2->1->4->3.

Note:

• Your algorithm should use only constant extra space.
• You may not modify the values in the list's nodes, only nodes itself may be changed.

蠢办法用n来标记有没有一对，

n = 1时移动到一对中的第一个，记录一下pre

n = 2时 要开始swap了，注意有3个next要变，一个是node.next = pre,pre.next = node.next, ppre.next = node可以画图理解一下，ppre是这一对之前的node，pre是这一对里第一个node，node是这一对里第二个node

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

class Solution:
    def swapPairs(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        n = 1
        if not head:return []
        if not head.next:return head
        else:cphead = head.next
        node = head
        ppre = None
        while(node):
            if n == 1:
                pre = node
                node = node.next
                n = 2
            elif n == 2:
                aft = node.next
                node.next = pre
                pre.next = aft
                if ppre != None:ppre.next = node
                n = 1
                ppre = pre
                node = aft
        return cphead