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描述
You are given a 0-indexed, strictly increasing integer array nums and a positive integer diff. A triplet (i, j, k) is an arithmetic triplet if the following conditions are met:
- i < j < k,
- nums[j] - nums[i] == diff, and
- nums[k] - nums[j] == diff.
Return the number of unique arithmetic triplets.
Example 1:
Input: nums = [0,1,4,6,7,10], diff = 3
Output: 2
Explanation:
(1, 2, 4) is an arithmetic triplet because both 7 - 4 == 3 and 4 - 1 == 3.
(2, 4, 5) is an arithmetic triplet because both 10 - 7 == 3 and 7 - 4 == 3.
Example 2:
Input: nums = [4,5,6,7,8,9], diff = 2
Output: 2
Explanation:
(0, 2, 4) is an arithmetic triplet because both 8 - 6 == 2 and 6 - 4 == 2.
(1, 3, 5) is an arithmetic triplet because both 9 - 7 == 2 and 7 - 5 == 2.
Note:
3 <= nums.length <= 200
0 <= nums[i] <= 200
1 <= diff <= 50
nums is strictly increasing.
解析
根据题意，给定一个 0 索引、严格递增的整数数组 nums 和一个正整数 diff 。 如果满足以下条件，则三元组 (i, j, k) 是算术三元组，返回所有不同的算术三元组的数量。：
- i < j < k
- nums[j] - nums[i] == diff
- nums[k] - nums[j] == diff
其实这道题目说过 nums 元素严格递增，这里面已经暗示了所有的 nums 元素都是不同的，然后我们只需要遍历 nums 中的每个元素 x ，如果 x-diff 存在于 nums 中并且 x+diff 存在于 nums 中，那就说明这个三元组是唯一的，结果 result 加一即可，遍历结束返回 result 。
时间复杂度为 O(N^2) ，空间复杂度为 O(1) 。
当然了，因为我们知道了 nums 中的元素都不同，为了提高速度，我们可以把 nums 变成集合，这样我们在查找元素的时候时间复杂度为 O(1) ，整体的时间复杂度变成了 O(N) 。
解答
class Solution(object):
def arithmeticTriplets(self, nums, diff):
"""
:type nums: List[int]
:type diff: int
:rtype: int
"""
result = 0
for x in nums:
if x-diff in nums and x+diff in nums:
result += 1
return result
运行结果
104 / 104 test cases passed.
Status: Accepted
Runtime: 50 ms
Memory Usage: 13.3 MB
原题链接
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