我创建了一个程序,使用分而治之算法计算数组的最大值,但输出为0。
#include <iostream>
using namespace std;
int array[50];
void maximum(int index1, int index2, int&max_number) {
int max_number1;
int max_number2;
int half;
if (index1 == index2)
max_number = array[index1];
else {
half = (index1 + index2) / 2;
maximum(index1, half, max_number1);
maximum(half + 1, index2, max_number2);
if(max_number1 < max_number2)
max_number = max_number2;
else
max_number = max_number1;
}
}
int main() {
int index1;
int index2;
int max_number = 0;
cout << "index2 = ";
cin >> index2;
for (index1 = 0; index1 < index2; index1++)
cin >> array[index1];
maximum(index1, index2, max_number);
cout << "maximum number = " << max_number;
return 0;
}
我该怎么办?
答案 0 :(得分:1)
此循环后
for(index1 = 0; index1 < index2; index1++)
cin>>array[index1];
index1
等于index2
。所以这个电话
maximum(index1, index2, max_number);
没有意义。
在任何情况下,函数都太复杂,并且使用全局变量,这是一种不好的编程习惯。
使用标准算法std::max
可以更简单地编写递归函数。
例如
#include <iostream>
#include <algorithm>
size_t maximum( const int a[], size_t n )
{
return n < 2 ?
0 :
std::max( maximum( a, n / 2 ),
n / 2 + maximum( a + n / 2, n - n / 2 ),
[a] ( size_t i, size_t j ) { return a[i] < a[j]; } );
}
int main()
{
const size_t N = 10;
int a[N] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
int b[N] = { 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 };
int c[N] = { 0, 2, 4, 6, 8, 9, 7, 5, 3, 1 };
size_t pos = maximum( a, N );
std::cout << "The maximum element is " << a[pos]
<< " at position " << pos << std::endl;
pos = maximum( b, N );
std::cout << "The maximum element is " << b[pos]
<< " at position " << pos << std::endl;
pos = maximum( c, N );
std::cout << "The maximum element is " << c[pos]
<< " at position " << pos << std::endl;
return 0;
}
程序输出
The maximum element is 9 at position 9
The maximum element is 9 at position 0
The maximum element is 9 at position 5
答案 1 :(得分:0)
你在这部分代码中有一个拼写错误
for(index1 = 0; index1 < index2; index1++)
index1将增加到index2,然后
maximum(index1, index2, max_number);
将在第1 /第2个参数中传递相同的值,您可以像
一样更改代码for(index1 = 0; index1 < index2; index1++)
cin>>array[index1];
index1 = 0;
maximum(index1, index2, max_number);