leetcode:Subsets

最新推荐文章于 2019-07-27 20:45:00 发布

原创最新推荐文章于 2019-07-27 20:45:00 发布 · 329 阅读

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本文探讨了如何求解一个给定整数集合的所有可能子集，并确保子集中元素顺序递增且不包含重复子集。通过具体示例说明了解决方案，例如对于集合[1,2,3]，输出包括所有可能的子集组合。

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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],
  []
]

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int getLastLetterIndex(vector<int> &S, vector<int> &curVector)
{
    int lastIndex = -1;
    if (curVector.size() == 0)
        return lastIndex;
    
    int lastLetter = curVector.at(curVector.size()-1);
    
    for (int i=0; i<S.size(); i++)
    {
        if (S.at(i) == lastLetter)
        {
            lastIndex = i;
            break;
        }
    }
    
    return lastIndex;
}


vector<vector<int> > subsets(vector<int> &S) {
    
    sort(S.begin(), S.end());
    
    vector<vector<int> > retVector;
    vector<int> curVector;
    
    retVector.push_back(curVector);
    
    int curIdx = 0;
    while (curIdx != retVector.size())
    {
        vector<int> curVector = retVector[curIdx];
        
        int loopIdx = getLastLetterIndex(S, curVector);
        
        for(int i=loopIdx+1; i<S.size(); i++)
        {
            vector<int> newVector = curVector;
            newVector.push_back(S.at(i));

            //print (newVector);
            retVector.push_back(newVector);
        }
        
        curIdx++;
    }
    
    return retVector;
}