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 == 0)   return NULL;
        ListNode *p1 = pListHead, *p2 = pListHead;
        int cnt = 1;
        while(cnt < k && p1!=NULL){
            p1 = p1->next;
            cnt = cnt+1;
        }
        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