zoukankan      html  css  js  c++  java
  • 0594. Longest Harmonious Subsequence (E)

    Longest Harmonious Subsequence (E)

    题目

    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

    题意

    在给定数组中找到一个子数组,使其中的最大值和最小值的差正好为1。

    思路

    根据题意,这样的子数组中只可能存在两种不同的数,有以下两种方法:

    先将数组排序,再利用滑动窗口找到首尾差为1的子数组,复杂度为(O(N))

    或可以用HashMap记录每个数字的出现次数,在遍历时,当前数为num,判断num-1和num+1是否存在,存在的话则比较max(ans, num出现次数 + (num-1) or (num+1)出现次数)。


    代码实现

    Java

    排序

    class Solution {
        public int findLHS(int[] nums) {
            if (nums.length == 1) {
                return 0;
            }
    
            int ans = 0;
            int left = 0, right = 1;
    
            Arrays.sort(nums);
            while (left <= nums.length - 2) {
                if (nums[right] - nums[left] == 1) {
                    ans = Math.max(ans, right - left + 1);
                    right++;
                } else if (nums[right] - nums[left] > 1) {
                    left++;
                } else {
                    right++;
                }
                if (left == right) right++;
                if (right == nums.length) break;
            }
    
            return ans;
        }
    }
    

    Hash

    class Solution {
        public int findLHS(int[] nums) {
            int ans = 0;
            Map<Integer, Integer> map = new HashMap<>();
            for (int num : nums) {
                int cnt = map.getOrDefault(num, 0) + 1;
                if (map.containsKey(num - 1)) {
                    ans = Math.max(ans, map.get(num - 1) + cnt);
                }
                if (map.containsKey(num + 1)) {
                    ans = Math.max(ans, map.get(num + 1) + cnt);
                }
                map.put(num, cnt);
            }
            return ans;
        }
    }
    
  • 相关阅读:
    linux安装redis 完整步骤
    java获取音频文件播放时长
    jar包部署在linux上后浏览器访问不到的问题
    FileRead方法
    FileWrite方法
    用Calendar方法知道月份的天数
    Calendar的用法
    两个时间相减(java简单用法)
    单列体现(Runtime)
    Random方法
  • 原文地址:https://www.cnblogs.com/mapoos/p/14374447.html
Copyright © 2011-2022 走看看