面试题22：链表中倒数第k个节点

 

# -*- 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
        # 先考虑k若大于链表的长度，返回None,链表的长度咋算，不太会,可以在遍历的时候count+1
        # 我的思路是，从头到尾遍历链表，将其存在list中，如果len(list)<k,返回None；如果k=0，也返回None
        
        result = []
        while head:
            result.append(head)
            head = head.next
        if k > len(result) or k == 0:
            return None
        return result[len(result)-k]

　思路2：参考书书中的解题思路，只遍历一遍链表，就可以找到倒数第k个节点。我们这样思考，倒数第k个就是正数第n-k+1个。可以设置2个指针，2个指针相差k-1步，当前面的走的指针走到最后了，那我们的后指针就走到了n-k+1的位置。

# -*- 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:
            return None
        pointer1 = head
        pointer2 = head
        count = 0
        
        while pointer1:
            if count > k-1:
                pointer2 = pointer2.next
            count += 1
            pointer1 = pointer1.next
        if count<k or k<=0:
            return None
        return pointer2