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 + diffk = j + diff if j in nums and k in nums:c +=1return c