《劍指offer》面試題22：鏈表中倒數第k個節點

題目描述
知識點：鏈表

使用前後指針即可，前指針先走k-1步，然後兩個指針一起移動，當前指針指向最後一個節點，後指針正好指向倒數第k個節點

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def FindKthToTail(self, head, k):
        # write code here
        if head == None or k <= 0:
            return None

        quick_pointer = head
        slow_pointer = head
        ans = []

        for i in range(k - 1):
            if quick_pointer.next:
                quick_pointer = quick_pointer.next
            else:
                return None

        while quick_pointer.next:
            quick_pointer = quick_pointer.next
            slow_pointer = slow_pointer.next
        return slow_pointer

拓展：若要求中間節點，可設置快慢指針，快指針一次走兩步，慢指針一次走一個，當快指針指向最後一個節點，慢指針指向中間節點。