1029. Median (25)

最新推荐文章于 2022-04-06 15:50:57 发布

原创最新推荐文章于 2022-04-06 15:50:57 发布 · 307 阅读

0 ·

CC 4.0 BY-SA版权

PAT 专栏收录该内容

22 篇文章

订阅专栏

本文介绍一种高效算法来找到两个已排序整数序列的中位数。通过合并两个序列并使用qsort进行排序，进而找出中位数。讨论了如何避免段错误，并通过直接合并和排序改进算法效率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

一连几题都不觉得难，让我产生了一种我也很强的错觉。。。。。

这题就是简单的排序，自从学会用qsort之后觉得排序问题都弱爆了。

一开始没有留意到题目给出的两个数组已经有序的，我就将两个数组合成一个，然后直接qsort。这时犯了一个逗比的错误，我知道两个1000000会超过数组最大长度，不能直接a[1000000]定义，会出现段错误。要用malloc申请空间，逗比的我只对sque这样做了，Q1和Q2继续直接定义，还思考了一阵为什么会出现段错误。

另外，看答案之后，发现有序之后，改写了一下，直接一边排序一边输入到第三个数组。结果比qsort快了近100ms

暴力qsort

merge

1029. Median (25)

时间限制
1000 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1={11, 12, 13, 14} is 12, and the median of S2={9, 10, 15, 16, 17} is 15. The median of two sequences is defined to be the median of the nondecreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.

Given two increasing sequences of integers, you are asked to find their median.

Input

Each input file contains one test case. Each case occupies 2 lines, each gives the information of a sequence. For each sequence, the first positive integer N (<=1000000) is the size of that sequence. Then N integers follow, separated by a space. It is guaranteed that all the integers are in the range of long int.

Output

For each test case you should output the median of the two given sequences in a line.

Sample Input
4 11 12 13 14
5 9 10 15 16 17
Sample Output
13

#include <stdio.h>
#include <stdlib.h>
#include<stdbool.h>

#define MAXLEN 2000010

typedef long int ElementType;

int myCmp(const void* a, const void* b)
{
    return *(int *)a - *(int *)b;
}

void testPrint(ElementType a[], int len)
{
    int i;
    for(i=0; i<len; i++)
    {
        printf("%ld ", a[i]);
    }
}

int main()
{

    int numOfQ1, numOfQ2;
    int total = 0;
    int i, j;

    //ElementType sque[MAXLEN];

    ElementType *sque = (ElementType *)malloc(sizeof(ElementType)*MAXLEN);

    scanf("%ld", &numOfQ1);
    //ElementType Q1[numOfQ1];
    ElementType *Q1 = (ElementType *)malloc(numOfQ1*sizeof(ElementType));
    for(i=0; i<numOfQ1; i++)
    {
        scanf("%ld", &Q1[i]);
        sque[total++] = Q1[i];
    }

    scanf("%ld", &numOfQ2);
    //ElementType Q2[numOfQ2];
    ElementType *Q2 = (ElementType *)malloc(numOfQ2*sizeof(ElementType));
    for(i=0; i<numOfQ2; i++)
    {
        scanf("%ld", &Q2[i]);
        sque[total++] = Q2[i];
    }

    qsort(sque, total, sizeof(sque[0]), myCmp);

//    i = j = 0;
//    while(i<numOfQ1 && j<numOfQ2)
//    {
//        if(Q1[i]<Q2[j])
//        {
//            sque[total++] = Q1[i++];
//        }
//        else
//        {
//            sque[total++] = Q2[j++];
//        }
//    }
//    while(i<numOfQ1)
//    {
//        sque[total++] = Q1[i++];
//    }
//    while(j<numOfQ2)
//    {
//        sque[total++] = Q2[j++];
//    }

    //testPrint(sque, total);

    int median;
    if(total&1)
    {
        median = total/2;
    }
    else
    {
        median = total/2 - 1;
    }
    printf("%ld", sque[median]);
    free(Q1);
    free(Q2);
    free(sque);
    system("pause");
}