【哈希表, 二分查找】算术三元组的数目

题目

6136. 算术三元组的数目

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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

方法一：哈希表

思路

把所有数存入哈希集合中，枚举所有数，如果集合中同时存在 num - diff 和 num + diff 就把计数器自增。

代码

class Solution {
    public int arithmeticTriplets(int[] nums, int diff) {
        Set<Integer> set = new HashSet<>(){{
            for (int num : nums) this.add(num);
        }};
        int count = 0;
        for (int num : nums) {
            if (set.contains(num - diff) && set.contains(num + diff)) count++;
        }
        return count;
    }
}

class Solution {
public:
    int arithmeticTriplets(vector<int>& nums, int diff) {
        unordered_set<int> st(nums.begin(), nums.end());
        int count = 0;
        for (int& num : nums) {
            if (st.count(num - diff) && st.count(num + diff)) ++count;
        }
        return count;
    }
};

方法二：二分查找

思路

数组是严格递增的，所以可以直接用二分查找 num - diff 和 num + diff ，如果同时存在直接把计数器自增。

代码

class Solution {
    public int arithmeticTriplets(int[] nums, int diff) {
        int count = 0;
        for (int num : nums) {
            if (Arrays.binarySearch(nums, num - diff) >= 0 &&
                Arrays.binarySearch(nums, num + diff) >= 0) count++;
        }
        return count;
    }
}

class Solution {
public:
    int arithmeticTriplets(vector<int>& nums, int diff) {
        int count = 0;
        for (int& num : nums) {
            if (binary_search(nums.begin(), nums.end(), num - diff) &&
                binary_search(nums.begin(), nums.end(), num + diff)) count++;
        }
        return count;
    }
};