1、题目

给你一个下标从 0 开始、严格递增 的整数数组 nums 和一个正整数 diff 。如果满足下述全部条件，则三元组 (i, j, k) 就是一个 算术三元组 ：

i < j < k ，
nums[j] - nums[i] == diff 且
nums[k] - nums[j] == diff
返回不同 算术三元组 的数目。

示例 1：

输入：nums = [0,1,4,6,7,10], diff = 3
输出：2
解释：
(1, 2, 4) 是算术三元组：7 - 4 == 3 且 4 - 1 == 3 。
(2, 4, 5) 是算术三元组：10 - 7 == 3 且 7 - 4 == 3 。
示例 2：

输入：nums = [4,5,6,7,8,9], diff = 2
输出：2
解释：
(0, 2, 4) 是算术三元组：8 - 6 == 2 且 6 - 4 == 2 。
(1, 3, 5) 是算术三元组：9 - 7 == 2 且 7 - 5 == 2 。

提示：

3 <= nums.length <= 200
0 <= nums[i] <= 200
1 <= diff <= 50
nums 严格 递增。

2、解

对nums进行遍历，每个三元组即为{num[i]. nums[i] + diff, nums[i] + 2*diff}
寻找其是否存在每个元素对应的三元组中的后两个元素, 如果存在结果+1。

    int arithmeticTriplets(vector<int> &nums, int diff){int result = 0;for(int i = 0; i < nums.size(); i++){vector<int>::iterator mid = find((nums.begin() + i), nums.end(), nums[i] + diff);vector<int>::iterator last = find((nums.begin() + i + 1), nums.end(), nums[i] + 2*diff);if(mid != nums.end() && last != nums.end()){result++;}}return result;}

也可以先将nums通过哈希集合存储后进行查找

    int arithmeticTriplets(vector<int>& nums, int diff) {unordered_set<int> hashSet;for (int x : nums) {hashSet.emplace(x);}int ans = 0;for (int x : nums) {if (hashSet.count(x + diff) && hashSet.count(x + 2 * diff)) {ans++;}}return ans;}