力扣78题

方法一： binary 不使用递归

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>> ans;
        int n = nums.size();
        for (int i = 0; i < 1 << n; i ++ ) {
            vector<int> path;
            for (int j = 0; j < n; j ++) {
                if (i >> j & 1) path.push_back(nums[j]);
            }
            ans.push_back(path);
        }
        return ans;
    }
};

方法二：DFS+backing

• 递归树很重要，之前以为简单，但是发现代码不会写才发现递归树已经蕴含了很多自己不懂的东西
• 
• 
• 

//DFS + back-tracing
//learn from huahua

class Solution {
private: 
    void dfs(vector<int>& nums, int n, int s, vector<int>& cur, vector<vector<int>>& ans) {
        if (n == cur.size()) {
            ans.push_back(cur);

            // <debug>
            cout << "ans is "<<' ';
            for (int i = 0; i < cur.size(); i ++ ) {
                cout << cur[i] <<' ';
            }
            cout << endl;
            // </debug>

            return ;
        }
        for (int i = s; i < nums.size(); ++ i) {
            cur.push_back(nums[i]);
            dfs(nums, n, i + 1, cur, ans);
            cur.pop_back();
        }
        // 错误代码
        /*
            这样的话 递归树就少了好几条路径, 还是要把递归树画出来才可以
            cur.push_back(nums[s]);
            dfs(nums, n, s + 1, cur, ans);
            cur.pop_back();
        */
    }
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>> ans;
        vector<int> cur;
        for (int i = 0; i <= nums.size(); i ++ ) {
            dfs(nums, i, 0, cur, ans);
        }
        return ans;
    }
};