We define a harmonious array is an array where the difference between its maximum value and its minimum value is exactly1.

Now, given an integer array, you need to find the length of its longest harmonious subsequence among all its possible .

Example 1:

Input: [1,3,2,2,5,2,3,7]Output: 5Explanation: The longest harmonious subsequence is [3,2,2,2,3].

 

Note: The length of the input array will not exceed 20,000.

 

Approach #1: C++.

class Solution {public:    int findLHS(vector
     
      & nums) {        int size = nums.size();        unordered_map
      
        temp;        for (int i = 0; i < size; ++i) {            temp[nums[i]]++;        }                vector
       
        
         > v(temp.begin(), temp.end());        sort(v.begin(), v.end(), cmp);                int ans = 0;        for (int i = 1; i < v.size(); ++i) {            if (v[i].first - v[i-1].first == 1)                 ans = max(ans, v[i].second + v[i-1].second);            //cout << ans << endl;        }                return ans;    }    private:    static bool cmp(pair
         
           a, pair
          
            b) { return a.first < b.first; }};
          
         
        
       
      
     

　　

Approach #2: Java.

class Solution {    public int findLHS(int[] nums) {        HashMap
     
       map = new HashMap<>();        int res = 0;        for (int num : nums) {            map.put(num, map.getOrDefault(num, 0) + 1);        }        for (int key : map.keySet()) {            if (map.containsKey(key + 1))                res = Math.max(res, map.get(key) + map.get(key + 1));        }        return res;    }}
     

　　

Approach #3: Python.

class Solution(object):    def findLHS(self, nums):        """        :type nums: List[int]        :rtype: int        """        count = collections.Counter(nums)        ans = 0        for x in count:            if x+1 in count:                ans = max(ans, count[x] + count[x+1])        return ans

　　

Time Submitted	Status	Runtime	Language
a few seconds ago		124 ms	python
3 minutes ago		68 ms	java
8 minutes ago		76 ms	cpp

Analysis:

At the first time I can't understand this question's mean, after listening to the explanation. I understand that this question let you find the subarray which the difference with the maximum and the minimum equal to 1. In the first approach I use a vector to sort the map and then calculate the sum elements which two elements have the difference equal to 1. but in the second approach I found we can don't use vector and sort because we can find the number which bigger one than other with the method of .count.