Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].

Return the answer in an array.
Example 1:
Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
Explanation: 
For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3). 
For nums[1]=1 does not exist any smaller number than it.
For nums[2]=2 there exist one smaller number than it (1). 
For nums[3]=2 there exist one smaller number than it (1). 
For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).

Example 2:
Input: nums = [6,5,4,8]
Output: [2,1,0,3]
Example 3:

Input: nums = [7,7,7,7]
Output: [0,0,0,0]
 
Constraints:
2 <= nums.length <= 500
0 <= nums[i] <= 100

方法一，首先尝试无脑暴力破解，先确保题目理解无误：

class Solution {
public:
    vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
        int n = nums.size();
        vector<int> res;
        for(int i = 0;i < n ;i++)
        {
            int count = 0;
            for(int j = 0;j < n ;j++)
            {
                if(nums[j] < nums[i])
                count++;
            }
            res.push_back(count);
        }
        
        return res;
    }
};

运行结果:

Runtime: 44 ms, faster than 46.05% of C++ online submissions for How Many Numbers Are Smaller Than the Current Number.
Memory Usage: 10.3 MB, less than 53.70% of C++ online submissions for How Many Numbers Are Smaller Than the Current Number.

优化解法时间复杂度O(n)，空间复杂度O(n)

class Solution {
public:
    vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
        int n = nums.size();
        vector<int> res(n,0);
        
        int freq[102] = {0};
        //we take nums[i] in nums[i] + 1 position,so get freq sum easy
        for(int i = 0;i < n;i++)
            freq[nums[i] + 1]++;
        
        for(int i = 1;i <102;i++)
            freq[i] += freq[i - 1];
        
        //get the res from freq array
        for(int i = 0;i < n;i++)
            res[i] = freq[nums[i]];
        
        return res;
    }
};

运行结果，击败99% C++提交

Runtime: 4 ms, faster than 99.70% of C++ online submissions for How Many Numbers Are Smaller Than the Current Number.
Memory Usage: 10.5 MB, less than 26.46% of C++ online submissions for How Many Numbers Are Smaller Than the Current Number.