You are given a 0-indexed, strictly increasing integer array  and a positive integer . A triplet  is an arithmetic triplet if the following conditions are met:

• i < j < k,

• nums[j] - nums[i] == diff, and

• nums[k] - nums[j] == diff.

Return the number of unique arithmetic triplets.

Example 1:

Input: 

nums = [0,1,4,6,7,10], 

diff = 3

Output: 2

Explanation:(1, 2, 4) is an arithmetic triplet because both 7 - 4 == 3 and 4 - 1 == 3. 

(2, 4, 5) is an arithmetic triplet because both 10 - 7 == 3 and 7 - 4 == 3.

Example 2:

Input: 

nums = [4,5,6,7,8,9], 

diff = 2

Output: 2

Explanation:(0, 2, 4) is an arithmetic triplet because both 8 - 6 == 2 and 6 - 4 == 2. 

(1, 3, 5) is an arithmetic triplet because both 9 - 7 == 2 and 7 - 5 == 2.

就是计算每个i开始能够形成多少等差数列，如果n>3，那么就可以形成n-2种不同的组合。

剩下就是计算多少个了，我用了一个boolean数组。去判断是否已经访问过了。

class Solution {
    public int arithmeticTriplets(int[] nums, int diff) {
         int n=nums.length;
        boolean[]flag=new boolean[n];
        Arrays.fill(flag, false);
        HashMap<Integer,Integer>map=new HashMap<>();
        for (int i = 0; i < n-1; i++) {
            if(!flag[i]){
                flag[i]=true;
                map.put(i, 1);
                int temp=1;
                for (int j = i+1; j < n; j++) {
                    if(nums[j]-nums[i]==temp*diff&&!flag[j]){
                        map.put(i, map.get(i)+1);
                        temp++;
                        flag[j]=true;
                    }
                }
            }
        }
        int cnt=0;
        for (Integer k : map.keySet()) {
            if(map.get(k)>=3){
                cnt+=map.get(k)-2;
            }
        }
        return cnt;
    }
}

Constraints:

• 3 <= nums.length <= 200

• 0 <= nums[i] <= 200

• 1 <= diff <= 50

• nums is strictly increasing.

Runtime: 3 ms, faster than 42.99% of Java online submissions for Number of Arithmetic Triplets.

Memory Usage: 40.6 MB, less than 52.22% of Java online submissions for Number of Arithmetic Triplets.