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法学C++ 程序寻找元素中值，这些元素存储在 2 个不同的数组中
我们考虑一个 C++ 程序，该程序寻找元素中值，其中元素存储在 2 个不同的数组中。
算法
Begin Function Median() has Two arrays a1[], a2[] and n = numbers of elements of the array as arguments: Initialize i and j by 0, and n1 and n2 by -1 for c in range 0 to n, do if i = n, then n1 := n2 n2 := a2[0] break the loop else if j = n, then n1 := n2 n2 := a1[0] break the loop if a1[i] < a2[j], then n1 := n2 n2 := a1[i] increase i by 1 else n1 := n2 n2 := a2[i] increase j by 1 done return the average of n1 and n2. End.
示例代码
#include<iostream>
#include <bits/stdc++.h>
using namespace std;
int Median(int a1[],int a2[], int n) {
int i = 0;
int j = 0;
int c;
int n1 = -1, n2 = -1;
for (c = 0; c <= n; c++) {
if (i == n) {
n1 = n2;
n2 = a2[0];
break;
}
else if (j == n) {
n1 = n2;
n2 = a1[0];
break;
}
if (a1[i] < a2[j]) {
n1 = n2;
n2 = a1[i];
i++;
} else {
n1 = n2;
n2 = a2[j];
j++;
}
}
return (n1 + n2)/2;
}
int main() {
int n1,n2, i;
cout<<"\nEnter the number of elements for 1st array: ";
cin>>n1;
int a1[n1];
for(i = 0; i < n1; i++) {
cout<<"Enter element for 1st array"<<i+1<<": ";
cin>>a1[i];
}
cout<<"\nEnter the number of elements for 2nd array: ";
cin>>n2;
int a2[n2];
for(i = 0; i < n2; i++) {
cout<<"Enter element for 2nd array "<<i+1<<": ";
cin>>a1[i];
}
if (n1 == n2)
cout << "Median is "
<< Median(a1, a2, n1) ;
else
cout << "Doesn't work for arrays"
<< " of unequal size";
return 0;
}输出
Enter the number of elements for 1st array: 5 Enter element for 1st array1: 2 Enter element for 1st array2: 4 Enter element for 1st array3: 6 Enter element for 1st array4: 7 Enter element for 1st array5: 9 Enter the number of elements for 2nd array: 5 Enter element for 2nd array 1: 20 Enter element for 2nd array 2: 40 Enter element for 2nd array 3: 60 Enter element for 2nd array 4: 70 Enter element for 2nd array 5: 90 Median is 20
更新于：30-7-2019
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