我想获得一个数组的所有组合使用C ++这样

I am trying to get all combinations of an array with C++ such that

double* list1 = new double[size];

在该数组的项目是 {1,2,3,4,5}

我需要添加所有可能的组合成一个堆栈，如：

I need to add all possible combinations into a stack, such as:

1+2, 1+2+3, 1+2+3+4，1+2+3+4+5, 1+3, 1+3+4, 1+3+4+5, 1+4, 1+4+5, 1+5...

我遇到的我做这通过2 for循环的问题，while循环

the problem I am running into is I am doing this through 2 for loops and a while loop

for(int i = 0; i < size - 1; i++)
{
    for(int j = i; j < size - 1; j++)
    {
        double temp = list1[i] + list1[j + 1];
        list1combos.push(temp);
        int k = j + 2;
        while (k <= size - 1)
        {
            temp = temp + list1[k];
            list1combos.push(temp);
            k++;
        }
    }
}

我可以得到那些我上面列出的，但我不知道如何code的组合，如 1 + 3 + 5 或 1 + 2 + 5

请就如何得到这些的组合，谢谢计算器！

Please advise on how to get those combinations, thanks stackoverflow!

推荐答案

由于顺序并不重要，我建议具有相同大小的数组作为你的 X 并在其上执行二进制增量，即你开始与inited仅 0 和数组数，直到你只有 1 秒。对于每一个另外的 1 你会选择置换从 X 阵列。

Since the order does not matter, I would suggest having an array of the same size as your x and perform a binary increment on it, i.e. you start with the array inited to only 0s and count until you have only 1s. For every addition of a 1 you would pick a permutation from your x array.

First iteration:
0 0 0 0 0 -> empty
Second iteration:
0 0 0 0 1 -> you pick 5
3rd iteration:
0 0 0 1 0 -> you pick 4
4th iteration:
0 0 0 1 1 -> you pick 4 and 5
And so on until you reach:
1 1 1 1 1 -> you pick 1, 2, 3, 4 and 5