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Leetcode:Subsets 子集生成

2024-07-08 16:30:55   215人阅读

Subsets：

Given a set of distinct integers, S, return all possible subsets.

Note:

  • Elements in a subset must be in non-descending order.
  • The solution set must not contain duplicate subsets.

 

For example,
If S = [1,2,3], a solution is:

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

 

解题分析：

每个元素，都有两种选择，选或者不选。

为了保持子集有序，必须首先对原数组sort

class Solution {public:    vector<vector<int> > subsets(vector<int> &S) {        vector<vector<int>> result;        if (S.size() == 0) return result;        sort(S.begin(), S.end());   // sort        vector<int> path;        dfs(S, path, 0, result);        return result;    }        void dfs(const vector<int>& S, vector<int>&path, int cur, vector<vector<int>>& result) {        if (cur == S.size()) {            result.push_back(path);            return;        }        // not choose S[cur]        dfs(S, path, cur + 1, result);        // choose S[cur]        path.push_back(S.at(cur));        dfs(S, path, cur + 1, result);        path.pop_back();    }};

 

 

Subsets II:

Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

  • Elements in a subset must be in non-descending order.
  • The solution set must not contain duplicate subsets.

 

For example,
If S = [1,2,2], a solution is:

[  [2],  [1],  [1,2,2],  [2,2],  [1,2],  []]

 

解题分析：

这题有重复元素，但本质上，跟上一题很类似，上一题中元素没有重复，相当于每个元素只能选0或1次，这里扩充到了每个元素可以选0到若干次而已

class Solution {public:    vector<vector<int> > subsetsWithDup(vector<int> &S) {        vector<vector<int>> result;        if (S.size() == 0) return result;        sort(S.begin(), S.end());        vector<int> path;        dfs(S, path, 0, result);        return result;    }        void dfs(const vector<int>& S, vector<int>& path, int cur, vector<vector<int>>& result) {        result.push_back(path);        for (int i = cur; i < S.size(); ++i) {            if (i != cur && S.at(i) == S.at(i-1)) {                continue;            }                        path.push_back(S.at(i));            dfs(S, path, i + 1, result);            path.pop_back();        }    }};

 

 


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