Leetcode篇：两两交换链表中的节点

Posted 2021-01-14 恩zzq我是

---

@author: ZZQ
@software: PyCharm
@file: swapPairs.py
@time: 2018/10/20 19:49
说明：给定一个链表，两两交换其中相邻的节点，并返回交换后的链表。
示例:
给定 1->2->3->4, 你应该返回 2->1->4->3.
说明:
你的算法只能使用常数的额外空间。
你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。
思路：
四个节点，分别记录当前需要进行交换的两个节点（first, second），以及这俩个节点的前后节点(pre, post)
然后每次只针对这四个节点进行交换即可。
注意考虑当输入是空节点，一个节点，两个节点，三个节点以及节点个数为奇数的情况。

class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution(object):
    def __init__(self):
        pass

    def exchange(self, pre, first, second, post):
        first.next = None
        second.next = None
        first.next = post
        second.next = first
        pre.next = second
    def swapPairs(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head is None or head.next is None:
            return head
        pre = ListNode(0)
        pre.next = head
        first = head
        second = head.next
        post = second.next
        p = pre
        while True:
            self.exchange(pre, first, second, post)
            pre = pre.next.next
            first = pre.next
            if first is None:
                break
            second = pre.next.next
            if second is None:
                break
            post = post.next.next
            if post is None:
                self.exchange(pre, first, second, post)
                break
        return p.next


if __name__ == "__main__":
    answer = Solution()
    l1 = ListNode(1)
    p1 = ListNode(2)
    p2 = ListNode(3)
    p3 = ListNode(4)
    p4 = ListNode(5)
    p5 = ListNode(6)
    l1.next = p1
    p1.next = p2
    p2.next = p3
    p3.next = p4
    p4.next = p5

    l2 = answer.swapPairs(l1)
    while l2 is not None:
        print l2.val
        l2 = l2.next