不将就就是发现的源动力——win8如何提升下载速度

Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.

For example,
If n = 4 and k = 2, a solution is:

[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

分析：

DFS + 剪枝。

class Solution {
public:
    vector<vector<int> > combine(int n, int k) {
        vector<int> path;
        vector<vector<int> > res;
        DFS(1, n, k, path, res);
        return res;
    }
private:
    void DFS(int cur, int n, int k, vector<int>&path, vector<vector<int> >&res)
    {
        // exit
        if(k == 0){
            res.push_back(path);
            return;
        }
        //only k elems left
        else if(n-cur+1 == k){
            for(int i=cur; i<=n; ++i)
                path.push_back(i);
            res.push_back(path);
            path.erase(path.end()-k, path.end());
            return;
        }
        //pick cur
        path.push_back(cur);
        DFS(cur+1, n, k-1, path, res);
        path.pop_back();
        
        // do not pick cur
        DFS(cur+1, n, k, path, res);
    }
};

不将就就是发现的源动力——win8如何提升下载速度,布布扣,bubuko.com

不将就就是发现的源动力——win8如何提升下载速度

原文：http://blog.csdn.net/u010926964/article/details/22800213