LintCode_249 Count of Smaller Number before itself

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 II and Count of Smaller Number before itself I first.

Have you met this question in a real interview? 

Yes

Example

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

思路是把数依次压入一个vector，不过在压入的时候找到该数的位置，维护这个vector的排序顺序：

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;    }};