题目1004：Median

最新推荐文章于 2021-10-21 14:47:55 发布

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本文介绍了一种算法，该算法能够处理两个已排序的整数序列，并找出这两个序列合并后的中位数。通过逐步合并两个序列并记录中间值的方式实现。适用于大数据集，且所有数值都在long int范围内。

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题目描述：

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 non-decreasing 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.

输入：

    Each input file may contain more than 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.

输出：

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

样例输入：

4 11 12 13 14
5 9 10 15 16 17

样例输出：

<span style="font-family:Courier New;">/*------------------------1004----------------------
description:Each input file may contain more than 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.
author     :yuanwei
time       :2013-02-12
--------------------------1004----------------------*/
 
#include <stdio.h>
#include <malloc.h>
 
//find the media number of the matrix a and the matrix b
int getMedia(long int *a, int lenOfA, long int *b, int lenOfB)
{
    int i = 0;
    int j = 0;
    int r = 0;
    long int *c = (long int *)malloc(sizeof(long int) * (lenOfA+lenOfB));
 
    if (NULL == c)
    {
        return 0;
    }
 
    while ((i<lenOfA) && (j<lenOfB))
    {
        if (*(a+i) < *(b+j))
        {
            *(c+r) = *(a+i);
            i++;
        }
        else
        {
            *(c+r) = *(b+j);
            j++;
        }
        r++;
    }
     
    while (i < lenOfA)
    {
        *(c+r) = *(a+i);
        r++;
        i++;
    }
    while (j < lenOfB)
    {
        *(c+r) = *(b+j);
        r++;
        j++;
    }
 
    i = (lenOfA+lenOfB)%2;
    if (i)
    {
        return *(c+(lenOfA+lenOfB)/2);
    }
    else
    {
        return *(c+(lenOfA+lenOfB)/2-1);
    }
 
}
 
int main()
{
    long int *a = NULL;
    long int *b = NULL;
    int lenOfA = 0;
    int lenOfB = 0;
    int media = 0;
    int i = 0;
 
    while (EOF != scanf("%d", &lenOfA))
    {
        a = (long int *)malloc(sizeof(long int) * lenOfA);
        if (NULL == a)
        {
            return 0;
        }
        for(i = 0; i < lenOfA; i++)
        {
            scanf("%d", &a[i]);
        }
         
        scanf("%d", &lenOfB);
        b = (long int *)malloc(sizeof(long int) * lenOfB);
        if (NULL == b)
        {
            return 0;
        }
        for(i = 0; i < lenOfB; i++)
        {
            scanf("%d", &b[i]);
        }
 
        media = getMedia(a, lenOfA, b, lenOfB);
        printf("%d\n", media);
 
        free(a);
        a = NULL;
        free(b);
        b = NULL;
         
    }
    return 0;
}
/**************************************************************
    Problem: 1004
    User: ruoshui1314
    Language: C++
    Result: Accepted
    Time:0 ms
    Memory:1020 kb
****************************************************************/</span>