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
点击查看代码
class Solution {
public int arithmeticTriplets(int[] nums, int diff) {
int len = nums.length,a =0;
for (int i = 0; i < len; i++) {
//二分查找，检索元素是否存在
int index1 = Arrays.binarySearch(nums,nums[i]+diff);
if (index1 > 0) {
//从检索元素的索引到结尾进行二分查找
int index2 = Arrays.binarySearch(nums,index1,len,nums[i]+2*diff);
if (index2 > 0) {
//记录算术三元组的数量
a++;
}
}
}
return a;
}
}