思路:暴力破解的时间复杂度为O(n2),且这里数组长度不大于500,也就是上万的数量级,肯定不会超时;
class Solution {
public:
vector<int> smallerNumbersThanCurrent(vector<int> &nums)
{
int len = nums.size();
vector<int> res(len, 0);
for (int i = 0; i < len; i++)
{
int temp = nums[i];
int count = 0;
for (int j = 0; j < len; j++)
{
if (nums[j] < temp)
{
count++;
}
}
res[i] = count;
}
return res;
}
};