We define a harmonious array as an array where the difference between its maximum value and its minimum value is exactly 1.

Given an integer array nums, return the length of its longest harmonious subsequence among all its possible subsequences.

A subsequence of array is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements.

 

Example 1:

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

Example 2:

Input: nums = [1,2,3,4]
Output: 2

Example 3:

Input: nums = [1,1,1,1]
Output: 0

 

Constraints:

• 1 <= nums.length <= 2 * 104
• -109 <= nums[i] <= 109

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

草草草草草 这么简单的

就是subsequence，不用连续那种，就用hashmap记录freq即可，然后对每个key找有没有key+1的key，有的话就比较再更新。