leetcode:2367. 算术三元组的数目(python3解法)

2023-12-13 05:27:02

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                  

文章来源:https://blog.csdn.net/qq_41905051/article/details/134948334
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