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。
OpenCV步步精深-可心科创工作室
思路：
既然diff已经给出来了，直接利用题目规则，j – diff在nums中且j + diff也在nums中就计数，否则不计数。
python3实现：
class Solution:
def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
count = 0
for j in nums:
if j - diff in nums and j + diff in nums:
count += 1
return count
OpenCV步步精深-可心科创工作室