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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:
Input: [5,2,6,1]
Output: [2,1,1,0]
Explanation:
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.
这道题我们主要参考了博文：here。我们从后往前遍历数据nums，并将每一个元素插入一个升序列表，这样该元素所在的位置的值就刚好是其右边小于该元素的元素个数。
class Solution {
public:
vector<int> countSmaller(vector<int>& nums) {
vector<int> record, res(nums.size());
for (int i = nums.size() - 1; i >= 0; i--) {
int d = distance(record.begin(), lower_bound(record.begin(), record.end(), nums[i]));
res[i] = d;
record.insert(record.begin() + d, nums[i]);
}
return res;
}
};
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