开启掘金成长之旅！这是我参与「掘金日新计划 · 2 月更文挑战」的第 8 天，点击查看活动详情

6136. 算术三元组的数目：

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 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。

样例 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 。

样例 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

分析

面对这道算法题目，二当家的陷入了沉思。
根据题意可以知道，当遍历到nums[k] 时，满足条件的三元组nums[j] = nums[k] - diff，nums[i] = nums[k] - diff - diff，只需要判断数字是否存在即可。

题解

rust

impl Solution {
    pub fn arithmetic_triplets(nums: Vec<i32>, diff: i32) -> i32 {
        let mut flag = vec![false; 201];

        nums.iter().filter(|&&n| {
            flag[n as usize] = true;
            n >= diff + diff && flag[(n - diff) as usize] && flag[(n - diff - diff) as usize]
        }).count() as i32
    }
}

go

func arithmeticTriplets(nums []int, diff int) int {
    ans := 0
    flag := make([]bool, 201)

    for _, n := range nums {
        flag[n] = true
        if n >= diff+diff && flag[n-diff] && flag[n-diff-diff] {
            ans++
        }
    }

    return ans
}

c++

class Solution {
public:
    int arithmeticTriplets(vector<int>& nums, int diff) {
        int ans = 0;
        bool flag[201] = {};

        for (int n: nums) {
            flag[n] = true;
            if (n >= diff + diff && flag[n - diff] && flag[n - diff - diff]) {
                ++ans;
            }
        }

        return ans;
    }
};

java

class Solution {
    public int arithmeticTriplets(int[] nums, int diff) {
        int ans = 0;
        boolean[] flag = new boolean[201];

        for (int n : nums) {
            flag[n] = true;
            if (n >= diff + diff && flag[n - diff] && flag[n - diff - diff]) {
                ++ans;
            }
        }

        return ans;
    }
}

python

class Solution:
    def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
        ans = 0
        flag = [False] * 201
        for n in nums:
            flag[n] = True
            if n >= diff + diff and flag[n - diff] and flag[n - diff - diff]:
                ans += 1
        return ans

非常感谢你阅读本文~
放弃不难，但坚持一定很酷~
希望我们大家都能每天进步一点点~
本文由二当家的白帽子：https://juejin.cn/user/2771185768884824/posts 博客原创~

开启掘金成长之旅！这是我参与「掘金日新计划 · 2 月更文挑战」的第 8 天，点击查看活动详情

【算法】6136. 算术三元组的数目（多语言实现）

6136. 算术三元组的数目：

样例 1：

样例 2：

提示：

分析

题解

rust

go

c++

java

python