7-4 The median of two ordered sequences (25 points)

7-4 The median of two ordered sequences (25 points)

Given that there are two equal-length non-descending sequences S1, S2, design the function to find the median of the union of S1 and S2. The median of the ordered sequence A​0​​,A​1​​,...,A​N−1​​refers to the value of A​(N−1)/2​​, that is, the first ⌊(N+1 )/2⌋ number (A​0​​is the first number).

Input format:

The input is divided into three lines. The first line gives the common length of the sequence N (0<N≤100000), and then each line enters the information of a sequence, that is, N integers in non-descending order. The numbers are separated by spaces.

Output format:

Output the median of the union sequence of the two input sequences in one line.

Input example 1:

5
1 3 5 7 9
2 3 4 5 6

 

Output sample 1:

4

 

Input example 2:

6
-100 -10 1 1 1 1
-50 0 2 3 4 5

 

Output sample 2:

1

Rearrange the two ordered sequences, (do not remove the duplicates, find the median);

 

Method 1: Input first, then compare and sort one by one 

 

#include <bits/stdc++.h>
using namespace std;

int main() {
	int n;
	cin>>n;
	vector<int>a,b,c;
	int x;
for(int i=0;i<n;++i)
{
	cin>>x;
	a.push_back(x);
}
for(int i=0;i<n;++i)
{
	cin>>x;
	b.push_back(x);
	
}
int i=0,j=0;
//把二者中小的一方放入数组c 
   while(!a.empty()&&!b.empty())
   {
   	if(a[i]>b[j])
   	{
   		c.push_back(b[j++]);
   		b.pop_back();
	   }
	   else
	   {
	   	c.push_back(a[i++]);
	   	a.pop_back();
	   }
   }
   if(!a.empty())
   {
   	while(!a.empty())
   	{
   		c.push_back(a[i++]);
   		a.pop_back();
	   }
   }
   if(!b.empty())
   {
   	while(!b.empty())
   	{
   		c.push_back(b[j++]);
   		b.pop_back();
	   }
   }
   int len=(c.size()-2)/2;
   cout<<c[len];
	return 0;
}

Method one can also put a counter when comparing the size, who is the n-1th number (the initial value of the counter is set to -1), who is the median

 

Method 2: Sort directly with sort 

#include <bits/stdc++.h>
using namespace std;

int main() {
	int n;
	cin>>n;
	vector<int>a;
	int x;
	for(int i=0;i<2*n;++i)
	{
		cin>>x;
		a.push_back(x); //vector不能直接使用cin>>a[i]; vecor只是声明，没有开辟内存 
	}
	sort(a.begin(),a.end());//对vecotr使用sort排序 
	cout<<a[n-1];//(2*n-2)/2=n-1

	return 0;
}