剑指offer：链表的中间节点

Posted 2021-10-21

给定一个链表的表头，如果链表节点数为奇数则返回中间的节点，否则返回中间两个节点的任意一个

# -*- coding: utf-8 -*-
# @Time         : 2019-04-23 17:24
# @Author       : Jayce Wong
# @ProjectName  : job
# @FileName     : findMidNode.py
# @Blog         : https://blog.51cto.com/jayce1111
# @Github       : https://github.com/SysuJayce

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

def FindKthToTail(head):
    """
    使用快慢指针，慢指针走1步，快指针走2步。当快指针指向尾节点的时候，慢指针所在的节点就是所求
    :param head: 
    :return: 
    """
    if not head:
        return None
    fast = slow = head
    while fast.next:
        fast = fast.next
        slow = slow.next
        if fast.next:
            fast = fast.next

    return slow

def main():
    zero = ListNode(0)
    one = ListNode(1)
    two = ListNode(2)
    three = ListNode(3)
    four = ListNode(4)

    zero.next = one
    one.next = two
    two.next = three
    three.next = four

    print(FindKthToTail(zero).val)

if __name__ == ‘__main__‘:
    main()