PAT 1029 Median [内存超限] [数组在线处理]

本文链接：https://blog.youkuaiyun.com/Time504/article/details/86772785

本文介绍了PAT 1029题目的解题思路和方法，主要问题是内存限制。解题关键是通过合并两个升序序列并排序找到中位数，但最初的数组存储方案会导致内存超限。解决方案是仅用变量存储一个序列，另一个序列按原样处理，从而减少内存使用。

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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 Specification:

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 (≤2×105) 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 Specification:

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:

--------------------------------------这是题目和解题的分割线--------------------------------------

这道题的思路很简单，就是排序取中位数，但是内存限制好坑的啊QAQ

我一开始的思路是用一个数组存取两个数组的内容，再用i和j两个指针模拟排序，可是最后一个测试点没法通过

//内存超限代码 
#include<cstdio>
#include<algorithm>
const int inf = 0x7fffffff;

using namespace std;

int n,i,j,m,a[400005];

int main()
{	
	scanf("%d",&n);
	for(i=0;i<n;i++)
		scanf("%d",&a[i]);
	scanf("%d",&m);
	for(i=n;i<m+n;i++)
		scanf("%d",&a[i]);
	i = j = 0;
	a[n+m] = inf;
	int mid = (m+n-1)/2;
	while(i+j<mid)
	{
		if(a[i]<a[n+j]) i++;
		else j++;
	}
	printf("%d",min(a[i],a[n+j]));
}

后来参考了柳神的代码，既然内存容易超限，那么可以只读取其中一个数组，另一个用普通变量读取即可。

//AC代码 
#include<cstdio> 
#include<algorithm>

//all the integers are in the range of long int.
//int类型的最大值为2^31-1，即0x7fffffff 
const int inf = 0x7fffffff;

using namespace std;

int n,i,j,m,tmp,a[200005];

int main()
{	
	scanf("%d",&n);
	for(i=0;i<n;i++)
		scanf("%d",&a[i]);
	scanf("%d",&m);
	a[n] = inf;
	//数组从0读取 
	int mid = (m+n-1)/2,x,count = 0;
	i = 0;
	//一边读取一边判断 
	while(m--)
	{
		scanf("%d",&tmp);
		//a数组更小的情况下，遍历 
		while(a[i]<tmp)	
		{	
			if(count==mid) printf("%d",a[i]);
			count++;i++;	
		} 				
		if(count==mid) printf("%d",tmp);
		count++;
	}
	//读取完了m个数还没找到中位数，即n远大于m的情况
	//这里的i接着上面的循环中i的位置 
	while(i<n)
	{
		if(count==mid) printf("%d",a[i]);
		count++;i++;
	}
}