leetcode:2367. 算术三元组的数目(python3解法）

难度：简单

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

题解：

class Solution(object):
    def arithmeticTriplets(self, nums, diff):
        for i in nums:
            j = i + diff
            k = j + diff 
            if j in nums and k in nums:
                c +=1
        return c