1534 Count Good Triplets
无脑上暴力破解解法：

class Solution {
public:
    int countGoodTriplets(vector<int>& arr, int a, int b, int c) {
        int res = 0,n = arr.size();
        for(int i = 0;i < n;i++)
        {
            for(int j = i + 1;j < n;j++)
            {
                for(int k = j + 1;k < n;k++)
                {
                    if(abs(arr[i] - arr[j]) <= a &&
                       abs(arr[j] - arr[k]) <= b &&
                       abs(arr[i] - arr[k]) <= c )
                        res++;
                }
            }
        }                 
        return res;
    }
};

运行结果正确，性能参数符合预期:

Runtime: 92 ms, faster than 5.15% of C++ online submissions for Count Good Triplets.
Memory Usage: 8.5 MB, less than 10.04% of C++ online submissions for Count Good Triplets.

对上面无脑暴力破解的算法进行优化，优化后的算法时间复杂度依然是n的三次方，运行结果击败80%用户。

Runtime: 32 ms, faster than 76.87% of C++ online submissions for Count Good Triplets.
Memory Usage: 8.5 MB, less than 9.74% of C++ online submissions for Count Good Triplets.

class Solution {
public:
    int countGoodTriplets(vector<int>& arr, int a, int b, int c) {
        int res = 0,n = arr.size();
        for(int i = 0;i < n - 2;i++)
        {
            for(int j = i + 1;j < n - 1;j++)
            {
                if (abs(arr[i] - arr[j]) > a )
                    continue;
                
                for(int k = j + 1;k < n;k++)
                {
                    if(abs(arr[j] - arr[k]) <= b &&
                       abs(arr[i] - arr[k]) <= c )
                        res++;
                }
            }
        }                 
        return res;
    }
};