牛客网－Ｃ++剑指offer－第十四题(链表中倒数第k个结点)

题目描述

输入一个链表，输出该链表中倒数第k个结点。

 

解题思路:这个题比较基础，主要考察链表的操作。两个思路:1.先将链表进行遍历，用vector将每个数据进行存储，然后直接找到某个数　２.先遍历链表，记录节点个数，再一次遍历，将数据取出

 

参考答案(时间复杂度大于O(n)):

#include <iostream>
#include <vector>
#include <string>
#include <stack>
#include <algorithm>
#include <math.h>

using namespace std;

struct ListNode {
    int val;
    struct ListNode *next;
};


class Solution {
public:
    ListNode* FindKthToTail(ListNode* pListHead, unsigned int k) {

        if (pListHead == NULL)
        {
            cout<<"pListHead = NULL fault"<<endl;
            return NULL;
        }


        int count = 1;
        ListNode* pmove = pListHead;

        while (pmove != NULL)
        {
            pmove = pmove->next;
            count++;
        }

        k = count - k;

        if (k <= 0)
        {
            cout<<"k <= 0 fault"<<endl;
            return NULL;
        }

        pmove = pListHead;
        for (int i = 0; i < k-1; ++i) {
            pmove = pmove->next;
        }

        return pmove;

    }
};

int main()
{
    //创建链表
    vector<int > array_list;

    Solution solution;

    //初始化链表内的参数
    ListNode ListNode3 = {3,NULL};
    ListNode ListNode2 = {2,&ListNode3};
    ListNode ListNode1 = {1,&ListNode2};

    //将头指针指向第一个链表的地址
    ListNode *head = &ListNode1;
    ListNode *my_result = NULL;

    my_result = solution.FindKthToTail(head,4);

    cout<<"result:"<<my_result->val<<endl;


    return 0;
}

 

优化代码:(使用vetocr进行存储，然后读取，时间复杂度为O(n))

class Solution {
public:
    ListNode* FindKthToTail(ListNode* pListHead, unsigned int k) {
        if(pListHead == NULL)
            return pListHead;

        vector<ListNode*> arr_l_node;
        ListNode* my_head = pListHead;
        while (my_head != NULL)
        {
            arr_l_node.push_back(my_head);
            my_head = my_head->next;
        }

        if (k > arr_l_node.size())
            return pListHead;

        return arr_l_node[arr_l_node.size()-k];
    }
};

 

更加优化代码(使用两个指针，时间复杂度同样为Ｏ(n),空间复杂度优于前面两种)

class Solution {
public:
    ListNode* FindKthToTail(ListNode* pListHead, unsigned int k) {
        //k如果等于0，因为是无符号数，k-1就变成了0xffffffff,而不是-1
        if(pListHead == NULL || k == 0)
            return pListHead;

        int count = 0;
        ListNode* P1;
        ListNode* P2;
        P1 = P2 = pListHead;

        //P2先走k-1步
        while (count < k-1)
        {
            //
            if (P2->next == NULL)
                return NULL;
            P2 = P2->next;
            count++;
        }

        //P1,P2一起走完
        while (P2->next != NULL)
        {
            P1 = P1->next;
            P2 = P2->next;
            count++;
        }

        return P1;
    }
};