子集生成相关问题

这里选取LeetCode两道子集生成的问题

• LC 78. 子集（集合中无重复元素)
• LC 90. 子集II（集合包含重复元素)

LC 78. 子集（集合中无重复元素）

DFS

const int N = 15;
class Solution {
public:
     bool vis[N];
     vector<vector<int>> ans;
     void dfs(vector<int>& nums, int x) {
        if (x == nums.size()) {
            vector<int> part;
            for (int i = 0; i < N; i ++ )
                if (vis[i]) part.push_back(nums[i]);
            ans.push_back(part);
            return ;
        }
        vis[x] = true;
        dfs(nums, x + 1);
        vis[x] = false;
        dfs(nums, x + 1);

     }
    vector<vector<int>> subsets(vector<int>& nums) {
        memset(vis, 0, sizeof vis);
        dfs(nums, 0);
        return ans;
    }
};

二进制枚举方法（推荐）

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

---

LC 90. 子集II（集合包含重复元素)

DFS

class Solution {
public:

    vector<vector<int>> ans;
    vector<int> part;
    void dfs(vector<int>& nums, int x) {
        if (x == nums.size()) {
            ans.push_back(part);
            return ;
        }
        int y = x + 1;
        while (y < nums.size() && nums[y] == nums[x]) y ++ ;
        for (int k = 0; k <= y - x; k ++ ) {
            dfs(nums, y);
            part.push_back(nums[x]);
        }

        for (int k = 0; k <= y - x; k ++ ) part.pop_back();
    }

    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        
        sort(nums.begin(), nums.end());
        dfs(nums, 0);
        
        return ans;
    }
};