leetcode: Subsets

最新推荐文章于 2020-06-13 14:11:24 发布

martin_liang

最新推荐文章于 2020-06-13 14:11:24 发布

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分类专栏： C++/C 算法 leetcode

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本文介绍了一种生成给定整数集合所有可能子集的算法。该算法确保子集中元素按非递减顺序排列且不含重复子集。通过递归方法实现，支持不同大小的输入集合。

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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],
  []
]

class Solution {
    void foo(int *array, int len, int remainNum, vector<int> &curVector, vector<vector<int>> &retVector)
    {   
        if (remainNum == 0)
        {
            retVector.push_back(curVector);
            return;
        }
           
        for (int i = 0; i<len; i++)
        {
            curVector.push_back(array[i]);
            foo(&array[i+1], len-i-1, remainNum-1, curVector, retVector);
            curVector.pop_back();
        }
    }
    
public:
    vector<vector<int> > subsets(vector<int> &S) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function  
        vector<vector<int>> retVector;
        vector<int> curVector;
        
        sort(S.begin(), S.end());
        
        retVector.push_back(curVector);

        int vectorSize = S.size();
        for (int subsetNum = 1; subsetNum <= vectorSize; subsetNum++)
        {
            for(int i = 0; i < vectorSize; i++)
            {
                curVector.push_back(S[i]);
                foo(&S[i+1], vectorSize-i-1, subsetNum-1, curVector, retVector);
                curVector.pop_back();
            }
        }   
        return retVector;
    }
};