我正在考虑网络中可用于我的优化项目的函数。我偶然发现了 Bateesh 提供的可用代码,我发现它非常有用。但我想稍微修改一下他的代码,这样
**不是打印输出,而是在我的主体中传输以用于下一个过程。
我该怎么做?我有新的 C++,所以任何建议都会有所帮助。我目前正在阅读有关传递数组的内容,但我对这些新想法感到茫然。
谢谢。
// Program to print all combination of size r in an array of size n
#include <stdio.h>
#include <stdlib.h>
void combinationUtil(int arr[], int n, int r, int count, int data[], int i);
// Needed for qsort. See http://w...content-available-to-author-only...s.com/reference/cstdlib/qsort/
int compare (const void * a, const void * b)
{
return ( *(int*)a - *(int*)b );
}
// The main function that prints all combinations of size r
// in arr[] of size n. This function mainly uses combinationUtil()
void printCombination(int arr[], int n, int r)
{
// A temporary array to store all combination one by one
int data[r];
// Sort array to handle duplicates
qsort (arr, n, sizeof(int), compare);
// Print all combination using temprary array 'data[]'
combinationUtil(arr, n, r, 0, data, 0);
}
/* arr[] ---> Input Array
n ---> Size of input array
r ---> Size of a combination to be printed
index ---> Current index in data[]
data[] ---> Temporary array to store current combination
i ---> index of current element in arr[] */
void combinationUtil(int arr[], int n, int r, int index, int data[], int i)
{
// Current cobination is ready, print it
if (index == r)
{
for (int j=0; j<r; j++)
printf("%d ",data[j]);
printf("\n");
return;
}
// When no more elements are there to be put
if (i >= n)
return;
// current is included, put next at next location
data[index] = arr[i];
combinationUtil(arr, n, r, index+1, data, i+1);
// Remove duplicates
while (arr[i] == arr[i+1])
i++;
// current is excluded, replace it with next (Note that
// i+1 is passed, but index is not changed)
combinationUtil(arr, n, r, index, data, i+1);
}
// Driver program to test above functions
int main()
{
int arr[] = {1, 2, 1, 3, 1};
int r = 3;
int n = sizeof(arr)/sizeof(arr[0]);
printCombination(arr, n, r);
return 0;
}
最佳答案
完全不要使用数组
我不会回答你问的问题,因为我认为原始数组在大多数情况下都很糟糕。如果它们被传递(这就是您所要求的),则更是如此。
但是,C++ 确实提供了原始数组的替代方案(想到 std::array
和 std::vector
)。除非您自己编写容器,否则我建议优先使用这些而不是原始数组。它们的性能非常可靠。
请使用 C++ 标准库函数
如果我没理解错的话,您是在尝试显示给定数字集合中唯一值的所有可能排列。您可以使用一些 C++ 库函数来执行此操作:
#include <algorithm>
#include <iostream>
#include <vector>
int main()
{
// Create vector
std::vector<int> v { 1, 2, 1, 3, 1 };
// Sort vector
std::sort(v.begin(), v.end());
// Move all duplicate entries to the end of the vector
auto it = std::unique(v.begin(), v.end());
// Trim vector so that the duplicates are no longer contained
v.resize(std::distance(v.begin(), it));
// Iterate as long as the function can rearrange the objects as a
// lexicographicaly greater permutation
do
{
// Print all elements in the vector
for(auto i : v)
std::cout << i << " ";
// Add new line character and flush output
std::cout << std::endl;
} while(std::next_permutation(v.begin(),v.end()));
return 0;
}
输出:
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
关于c++ - 将数组传回主体,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44983257/