LintCode_249 Count of Smaller Number before itself

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本文介绍了一种不使用线段树实现的计数较小数算法。该算法通过遍历数组并利用lower_bound函数来查找小于当前元素的数量，最终返回一个记录了每个元素前面有多少个比它小的数的数组。

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Give you an integer array (index from 0 to n-1, where n is the size of this array, value from 0 to 10000) . For each element Ai in the array, count the number of element before this element Ai is smaller than it and return count number array.

Notice

We suggest you finish problem Segment Tree Build, Segment Tree Query IIand Count of Smaller Number before itself I first.

Example

For array [1,2,7,8,5], return [0,1,2,3,2]

不用线段树，竟然也过了~~~

class Solution {
public:
   /**
     * @param A: An integer array
     * @return: Count the number of element before this element 'ai' is 
     *          smaller than it and return count number array
     */
    vector<int> countOfSmallerNumberII(vector<int> &A) {
        // write your code here
        vector <int> temp;
        vector<int> res;
        for (int i = 0; i < A.size(); i++) {
            auto itr = lower_bound(temp.begin(), temp.end(), A[i]);
            int count = itr - temp.begin();
            res.push_back(count);
            temp.insert(itr, A[i]);
        }
        return res;
    }
};