【创作赢红包】2367. 算术三元组的数目

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

方法一：暴力枚举
为了得到算术三元组的数目，最直观的做法是使用三重循环暴力枚举数组中的每个三元组，判断每个三元组是否为算术三元组，枚举结束之后即可得到算术三元组的数目。

class Solution {
    public int arithmeticTriplets(int[] nums, int diff) {
        int ans = 0;
        int n = nums.length;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (nums[j] - nums[i] != diff) {
                    continue;
                }
                for (int k = j + 1; k < n; k++) {
                    if (nums[k] - nums[j] == diff) {
                        ans++;
                    }
                }
            }
        }
        return ans;
    }
}

方法二：哈希集合

class Solution {
    public int arithmeticTriplets(int[] nums, int diff) {
        Set set = new HashSet();
        for (int x : nums) {
            set.add(x);
        }
        int ans = 0;
        for (int x : nums) {
            if (set.contains(x + diff) && set.contains(x + 2 * diff)) {
                ans++;
            }
        }
        return ans;
    }
}