Careercup

2014-05-02 02:28

题目链接

原题：

I/P: N, k 

O/P: all subset of N with exactly K elements. 

eg: I/p: N = 5, K =3 
O/p: 

1 2 3 
1 2 4 
1 2 5 
1 3 4 
1 3 5 
2 3 4 
2 3 5 
3 4 5

题目：从N个数中选K个，输出所有选法。

解法：递归解决。

代码：

// http://www.careercup.com/question?id=6204973461274624
#include <vector>
using namespace std;

class Solution {
public:
    void findKSubsets(int n, int k, vector<vector<int> > &result) {
        if (n <= 0) {
            return;
        }
        if (k < 1 || k > n) {
            return;
        }
        
        vector<int> subset;
        
        DFS(0, 0, subset, result, n, k);
    }
private:
    void DFS(int in, int ik, vector<int> &subset, 
             vector<vector<int> > &result, 
             const int &n, const int &k) {
        if (ik == k) {
            result.push_back(subset);
            return;
        }
        
        int i;
        for (i = in + 1; i <= n - (k - ik) + 1; ++i) {
            subset.push_back(i);
            DFS(i, ik + 1, subset, result, n, k);
            subset.pop_back();
        }
    };
};

int main()
{
    vector<vector<int> > result;
    int n, k;
    int i, j;
    Solution sol;
    
    while (scanf("%d%d", &n, &k) == 2 && (n > 0 && k > 0)) {
        sol.findKSubsets(n, k, result);
        printf("{
");
        for (i = 0; i < (int)result.size(); ++i) {
            printf("    {");
            for (j = 0; j < (int)result[i].size(); ++j) {
                printf((j ? ", %d" : "%d"), result[i][j]);
            }
            printf("}
");
        }
        printf("}
");
        
        for (i = 0; i < (int)result.size(); ++i) {
            result[i].clear();
        }
        result.clear();
    }
    
    return 0;
}