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LintCode "Count of Smaller Number"

时间:2015-10-01 11:32:16      阅读:111      评论:0      收藏:0      [点我收藏+]

Fenwick Tree is perfect for this problem, though space complexity is not quite efficient.

class Solution {
    //////////////////
    // Fenwick Tree //
    vector<int> ft;
    void update(int i, int x)
    {
        if((i + 1) > ft.size()) return;
        for (; i < ft.size(); i += (i & -i))
            ft[i] += x;
    }
    
    int query(int i)
    { 
        i = min(i, int(ft.size() - 1));
        int s = 0;  
        for (; i > 0; i -= (i & -i)) 
            s += ft[i];  
        return s; 
    }
    //////////////////
public:
   /**
     * @param A: An integer array
     * @return: The number of element in the array that 
     *          are smaller that the given integer
     */
    vector<int> countOfSmallerNumber(vector<int> &A, vector<int> &queries) {
        int maxNo = 0;
        if(A.size() > 0)
        {
            maxNo = *max_element(A.begin(), A.end());
        }
        ft.assign(maxNo + 1, 0); // memory complexity may not be good
        
        for(auto v : A)
        {
            update(v, 1);
        }
        
        vector<int> ret;
        for(auto i : queries)
        {
            int r = query(i - 1);
            ret.push_back(r);
        }
        return ret;
    }
};

LintCode "Count of Smaller Number"

原文:http://www.cnblogs.com/tonix/p/4850797.html

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