Count of Smaller Numbers After Self

Count of Smaller Numbers After Self

Description:

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example:

Given nums = [5, 2, 6, 1]

To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.
Return the array [2, 1, 1, 0].

分析：

这道题与Reverse Pairs基本一样，知识需要一个数组来做记录，详细细节可以看那里。

代码如下：

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class Solution {
public:
    vector<int> countSmaller(vector<int>& nums) {
        vector<int> t, res(nums.size());
        for (int i = nums.size() - 1; i >= 0; --i) {
            int left = 0, right = t.size();
            while (left < right) {
                int mid = left + (right - left) / 2;
                if (t[mid] >= nums[i]) right = mid;
                else left = mid + 1;
            }
            res[i] = right;
            t.insert(t.begin() + right, nums[i]);
        }
        return res;
    }
};