算法leetcode｜6136. 算术三元组的数目（rust和go全部双百）

Posted 2022-10-07 二当家的白帽子

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文章目录

• 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
• 原题传送门：https://leetcode.cn/problems/number-of-arithmetic-triplets/

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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

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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。
  

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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
    

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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

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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;
    
;

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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;
    

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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

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原题传送门：https://leetcode.cn/problems/number-of-arithmetic-triplets/

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