leetcode 594. Longest Harmonious Subsequence

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

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

Example 1:

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

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

本质上是一个组合问题，C（n，2），但是排序可以降低下复杂度：

class Solution(object):
    def findLHS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        # brute force
        cnt = collections.Counter(nums)
        items = cnt.items()
        items.sort(key=lambda x:x[0])
        ans = 0       
        for i in xrange(1, len(items)):
            k,v = items[i-1]：
            k2,v2 = items[i]
            if k2-k == 1:
                ans = max(ans, v2+v)
        return ans                        

还可以简化下：

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

此外，可以使用排序，本质上是自己build一个hash：

class Solution(object):
    def findLHS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        # brute force
        nums.sort()
        pre_k = pre_cnt = None
        ans = 0
        i = 0
        while i < len(nums):
            k = nums[i]
            cnt = 1
            while i+1<len(nums) and nums[i+1] == k:
                i += 1
                cnt += 1
            if pre_k is not None:           
                if k-pre_k==1: ans = max(pre_cnt+cnt, ans)                 
            pre_k = k
            pre_cnt = cnt            
            i += 1
        return ans     
        

其他解法：

int findLHS(vector<int>& nums) {
       sort(nums.begin(),nums.end());
       int len = 0;
       for(int i = 1, start = 0, new_start = 0; i<nums.size(); i++)
       {
 
           if (nums[i] - nums[start] > 1)    
               start = new_start;
           if (nums[i] != nums[i-1]) 
               new_start = i;
           if(nums[i] - nums[start] == 1)
               len = max(len, i-start+1);
       }
       return len;

另外，如果是连续的子序列，则本质上是一个计数问题：

class Solution(object):
    def findLHS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        # if sub seq is continueous
        if not nums: return 0
        min_val = max_val = nums[0]
        ans = cnt = 1
        for i in xrange(1, len(nums)):
            max_val = max(nums[i], max_val)
            min_val = min(nums[i], min_val)
            if max_val - min_val <= 1:                
                cnt += 1
                ans = max(ans, cnt)
            else:
                min_val = max_val = nums[i]
                cnt = 1
        return ans                