LeetCode（Map)2367. Number of Arithmetic Triplets

小鱼儿 2023-09-24 17:09 26阅读 0赞

# 1.问题 #

You are given a 0-indexed, strictly increasing integer array nums and a positive integer diff. A triplet (i, j, k) is an arithmetic triplet if the following conditions are met:

i < j < k,  
nums\[j\] - nums\[i\] == diff, and  
nums\[k\] - nums\[j\] == diff.  
Return the number of unique arithmetic triplets.

Example 1:

> Input: nums = \[0,1,4,6,7,10\], diff = 3  
> Output: 2  
> Explanation:  
> (1, 2, 4) is an arithmetic triplet because both 7 - 4 == 3 and 4 - 1 == 3.  
> (2, 4, 5) is an arithmetic triplet because both 10 - 7 == 3 and 7 - 4 == 3.

Example 2:

> Input: nums = \[4,5,6,7,8,9\], diff = 2  
> Output: 2  
> Explanation:  
> (0, 2, 4) is an arithmetic triplet because both 8 - 6 == 2 and 6 - 4 == 2.  
> (1, 3, 5) is an arithmetic triplet because both 9 - 7 == 2 and 7 - 5 == 2.

Constraints:

*  3 <= nums.length <= 200
 *  0 <= nums\[i\] <= 200
 *  1 <= diff <= 50
 *  nums is strictly increasing.

# 2. 解题思路 #

## 方法1： ##

> 1.新建一个HashSet为set，通过for循环遍历数组添加到set中  
> 2.for循环遍历，set是否包含nums数组中，元素+diff,元素+2\*diff，如果有result加1  
> 3.返回result

## 方法2： ##

> 暴力解法，可以参考

# 3. 代码 #

## 代码1： ##

class Solution {
        
        public int arithmeticTriplets(int[] nums, int diff) {
        
            int result = 0;
            Set<Integer> set = new HashSet<>();//1.新建一个HashSet为set，通过for循环遍历数组添加到set中
            for(int n:nums){
        
                set.add(n);
            }
            for(int s:nums){
        //2.for循环遍历，set是否包含nums数组元素，元素+diff,元素+2*diff，如果有result加1
                if(set.contains(s+diff)&&set.contains(s+2*diff)){
        
                    result+=1;
                }
            }
            return result;//3.返回result
            
        }
    }

> 解题思路基本相同

public int arithmeticTriplets(int[] nums, int diff) {
        
            int result = 0;
            Set<Integer> set = new HashSet<>();
            for (int num : nums) {
        //for循环遍历，如果set中没有数据，set.add（）添加数组元素。如果set中存在num - diff&& num - diff * 2，result+1
                if (set.contains(num - diff) && set.contains(num - diff * 2))
                    result++;
    
                set.add(num);
            }
            return result;
        }

> 解题思路基本，使用hashmap

class Solution {
        
        public int arithmeticTriplets(int[] nums, int diff) {
        
            int result = 0;
            HashMap<Integer, Integer> map = new HashMap<>();
            for(int i : nums){
        //1.新建一个hashmap，通过for循环遍历数组添加到map中，map的key为nums，value为1
                map.put(i, map.getOrDefault(i, 0) + 1);
            }
            for (int num : nums) {
        //2.for循环遍历，map的key是否包含nums数组元素(元素+diff,元素+2*diff),如果有result加1
                if (map.containsKey(num+diff) && map.containsKey(num + diff * 2))
                    result++;
            }
            return result;//3.返回result
            
        }
    }

## 代码2： ##

public int arithmeticTriplets(int[] nums, int diff) {
        
            int count = 0;
    
            for (int i = 0; i < nums.length - 2; i++) {
        
                for (int j = i + 1; j < nums.length - 1; j++) {
        
                    for (int k = j + 1; k < nums.length; k++) {
        
                        if (nums[j] - nums[i] == diff && nums[k] - nums[j] == diff)
                            count++;
                    }
                }
            }
            return count;
        }