剑指offer：链表中倒数第k个结点

问题描述

输入一个链表，输出该链表中倒数第k个结点。

解题思路

两个指针都指向头结点，第一个指针先移动k-1个结点，之后两指针同时移动，当第一个指针到链表尾的时候，第二个指针刚好指向倒数第k个结点。

c++代码

 /*
 struct ListNode {
     int val;
     struct ListNode *next;
     ListNode(int x) :
             val(x), next(NULL) {
     }
 };*/
 class Solution {
 public:
     ListNode* FindKthToTail(ListNode* pListHead, unsigned int k) {
         if(pListHead == NULL || k == )   return NULL;
         ListNode *p1 = pListHead, *p2 = pListHead;
         int cnt = ;
         while(cnt < k && p1!=NULL){
             p1 = p1->next;
             cnt = cnt+;
         }
         if(cnt < k || p1 == NULL)  return NULL;
         while(p1->next != NULL){
             p1 = p1->next;
             p2 = p2->next;
         }
         return p2;
     }
 };

python代码

 # -*- 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 is None or k == 0:
             return None
         p1, p2 = head, head
         cnt = 1
         while cnt < k and p1:
             p1 = p1.next
             cnt = cnt + 1
         if cnt < k or p1 is None:
             return None
         while p1.next:
             p1 = p1.next;
             p2 = p2.next;
         return p2