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LeetCode 594. Longest Harmonious Subsequence

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题目

原题在此
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 * $10^4$
  • $-10^9$ <= nums[i] <= $10^9$

解析

Solution_0

将每个元素的出现次数记录在map里. 因为map是sorted order, 遍历map 更新最大值就好.

Solution_1

排序后处理, 在每次数字变了的时候重新计算最大值. (虽然排序是$O(logN)$, 但因为是built_in反而更快, 且空间复杂度是$O(1)$.)

代码

Solution_0

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class Solution {
public:
    int findLHS(vector<int>& nums) {
        unordered_map<int,int>mp;
        for(auto i: nums)
            mp[i]++;
        int res = 0;
        for (auto [key, val] : mp)
            if(mp.count(key + 1))
                res = max(res, val + mp[key + 1]);
        return res;
    }
};

Solution_1

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class Solution {
public:
    int findLHS(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int res = 0, n = nums.size();
        int i = 0, j = 0;
        while (j < nums.size()) {
            while (j < n && (nums[j] - nums[i] <= 1)) j++;
            res = max(res, (nums[j - 1] == nums[i]) ? 0 : (j - i));
            if(j == n) break;
            while (i < n && (nums[j] - nums[i] >  1)) i++;
        }
        return res;
    }
};