目录链接:
力扣编程题-解法汇总_分享+记录-CSDN博客
GitHub同步刷题项目:
https://github.com/September26/java-algorithms
原题链接:力扣
描述:
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
解题思路:
* 解题思路: * 把nums转换为set,然后遍历nums,判断是否存在num + diff和num + 2 * diff的数即可。
代码:
public class Solution2367 {
public int arithmeticTriplets(int[] nums, int diff) {
Set<Integer> set = Arrays.stream(nums).boxed().collect(Collectors.toSet());
int abs = 0;
for (int i = 0; i < nums.length; i++) {
int num = nums[i];
if (set.contains(num + diff) && set.contains(num + 2 * diff)) {
abs++;
}
}
return abs;
}
}