PAT (Advanced Level) Practice — 1029 Median （25 分）_pat (advanced level) practice 1029 median-优快云博客

本文介绍了一种高效算法，用于求解两个已排序整数序列的中位数，避免了直接合并排序带来的内存超限问题。通过在线处理第二个序列，逐个比较并更新计数，确保了算法的时间和空间效率。

题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805466364755968

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:

：题意 给两个已经排好序的数列，求合并这两个序列后的中位数。

不能简单地输入两个数列，然后排序，这样会内存超限

先输入一个数列的值，将另一个数列的值在线处理，每输入一个值，循环判断如果前面数组的值小于第二个数组的值，ans++，cnt(第一个数组)++，直到ans到达中位数的位置（n+m+1）/2，
如果这样数完还没有到达mid，那么ans++，一直到等于mid，输出当前第二个数组输入的值。
如果第二个数组数完了还没到达mid，那么ans++，一直到mid，输出第一个数组的cnt位置上的值。

#include<iostream>
#include<algorithm>
using namespace std;
const int maxx=100010;
int a[2*maxx];
int main(){
	int n,m;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
	}
	a[n+1]=0x7fffffff;
	scanf("%d",&m);
	int t;
	int cnt=1;
	int ans=0;
	int mid=(n+m+1)/2;
	for(int i=1;i<=m;i++){
		scanf("%d",&t);
		while(a[cnt]<t){
			ans++;
			if(ans==mid){
				cout<<a[cnt]<<endl; 
				break;
			}
			cnt++;
		}
		ans++;
		if(ans==mid){
			cout<<t<<endl;
			break;
		} 
	}
	while(cnt<=n){
		ans++;
		if(ans==mid){
			cout<<a[cnt]<<endl;
			break;
		}
		cnt++;
	}
	return 0;
}