迭代任何大小的子集

Question

我可以遍历大小为1的子集

for( int a = 0; a < size; a++ ) {

或者大小为2的子集

for( int a1 = 0; a1 < size; a1++ ) {
    for( int a2 = a1+1; a2 < size; a2++ ) {

或3

for( int a1 = 0; a1 < size; a1++ ) {
for( int a2 = a1+1; a2 < size; a2++ ) {
   for( int a3 = a2+1; a3 < size; a3++ ) {

但是如何对大小为n的子集执行此操作？

根据Adam Rosenfield的回答,这可以完成这项工作

void iterate(int *a, int i, int size, int n)
{
 int start = 0;
 if( i > 0 ) start = a[i-1]+1;
 for(a[i] = start; a[i] < n; a[i]++) {
  if(i == n-1) {
      // a is the array of indices of size n
      for( int k = 0; k < size; k++ ) {
          printf("%d ",a[k]);
      }
      printf("\n");
  }
        else
            iterate(a, i+1, size, n);
    }
}

Answer 1

你可以使用递归:

void iterate(int *a, int i, int size, int n)
{
    for(a[i] = 0; a[i] < size; a[i]++)
    {
        if(i == n-1)
            DoStuff(a, n);  // a is the array of indices of size n
        else
            iterate(a, i+1, size, n);
    }
}
...
// Equivalent to 4 nested for loops
int a[4];
iterate(a, 0, size, 4);