LC.P2367[算数三元组的数目]

题目描述

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

方法一：暴力

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class Solution {
    public int arithmeticTriplets(int[] nums, int diff) {
        int n = nums.length, ans = 0;
        for (int j = 1; j < n - 1; ++j) {
            for (int i = 0; i < j; ++i) {
                for (int k = j + 1; k < n; ++k) {
                    if (nums[j] - nums[i] == diff && nums[k] - nums[j] == diff) ++ans;
                }
            }
        }
        return ans;
    }
}
  • 时间复杂度：$O(n^3)$
  • 空间复杂度：$O(1)$

方法二：哈希表

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class Solution {
    public int arithmeticTriplets(int[] nums, int diff) {
        int ans = 0;
        boolean[] visited = new boolean[301];
        for (int num : nums) {
            visited[num] = true;
        }
        for (int num : nums) {
            if (visited[num + diff] && visited[num + 2 * diff]) ++ans;
        }
        return ans;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$