Combinations

描述

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],
]

递归

// Combinations
// 深搜,递归
// 时间复杂度O(n!),空间复杂度O(n)
class Solution {
public:
    vector<vector<int> > combine(int n, int k) {
        vector<vector<int> > result;
        vector<int> path;
        dfs(n, k, 1, 0, path, result);
        return result;
    }
private:
    // start,开始的数, cur,已经选择的数目
    static void dfs(int n, int k, int start, int cur,
            vector<int> &path, vector<vector<int> > &result) {
        if (cur == k) {
            result.push_back(path);
        }
        for (int i = start; i <= n; ++i) {
            path.push_back(i);
            dfs(n, k, i + 1, cur + 1, path, result);
            path.pop_back();
        }
    }
};

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