Count of Smaller Numbers After Self
Desicription
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:
代码语言：javascript复制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.Solution
代码语言：javascript复制class Solution {
public:
std::vector<int> countSmaller(std::vector<int>& nums) {
auto newNums = std::vector<int>();
auto res = std::vector<int>(nums.size());
for(int i = nums.size() - 1; i >= 0; i--) {
auto distance = std::distance(newNums.begin(), std::lower_bound(newNums.begin(), newNums.end(), nums[i]));
res[i] = distance;
newNums.insert(newNums.begin() distance, nums[i]);
}
return res;
}
};
