LeetCode 2367.等差三元组的数目

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 等差三元组 ：

i < j < k ，
nums[j] - nums[i] == diff 且
nums[k] - nums[j] == diff
返回不同 等差三元组 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) 是等差三元组：7 - 4 == 3 且 4 - 1 == 3 。
(2, 4, 5) 是等差三元组：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) 是等差三元组：8 - 6 == 2 且 6 - 4 == 2 。
(1, 3, 5) 是等差三元组：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

三指针模拟即可：

class Solution {
public:int arithmeticTriplets(vector<int>& nums, int diff) {int i = 0;int j = 1;int k = 2;int ans = 0;while (k < nums.size()) {if (nums[k] - nums[j] < diff) {++k;continue;}if (nums[j] - nums[i] < diff) {++j;k = max(j + 1, k);continue;}if (nums[j] - nums[i] == diff && nums[k] - nums[j] == diff) {++ans;}++i;j = max(i + 1, j);k = max(j + 1, k);}return ans;}
};

如果nums的长度为n，则此算法时间复杂度为O(n)，空间复杂度为O(1)。