一個小小工程師的心情抒發天地

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]

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

Constraints:

• 2 <= nums.length <= 500
• 0 <= nums[i] <= 100