• [Swift]LeetCode1121. 将数组分成几个递增序列 | Divide Array Into Increasing Sequences


    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
    ➤微信公众号:山青咏芝(shanqingyongzhi)
    ➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/
    ➤GitHub地址:https://github.com/strengthen/LeetCode
    ➤原文地址:https://www.cnblogs.com/strengthen/p/11179565.html 
    ➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
    ➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

    Given a non-decreasing array of positive integers nums and an integer K, find out if this array can be divided into one or more disjoint increasing subsequences of length at least K

    Example 1:

    Input: nums = [1,2,2,3,3,4,4], K = 3
    Output: true
    Explanation: 
    The array can be divided into the two subsequences [1,2,3,4] and [2,3,4] with lengths at least 3 each.
    

    Example 2:

    Input: nums = [5,6,6,7,8], K = 3
    Output: false
    Explanation: 
    There is no way to divide the array using the conditions required. 

    Note:

    1. 1 <= nums.length <= 10^5
    2. 1 <= K <= nums.length
    3. 1 <= nums[i] <= 10^5

    给你一个 非递减 的正整数数组 nums 和整数 K,判断该数组是否可以被分成一个或几个 长度至少 为 K 的 不相交的递增子序列。 

    示例 1:

    输入:nums = [1,2,2,3,3,4,4], K = 3
    输出:true
    解释:
    该数组可以分成两个子序列 [1,2,3,4] 和 [2,3,4],每个子序列的长度都至少是 3。
    

    示例 2:

    输入:nums = [5,6,6,7,8], K = 3
    输出:false
    解释:
    没有办法根据条件来划分数组。 

    提示:

    1. 1 <= nums.length <= 10^5
    2. 1 <= K <= nums.length
    3. 1 <= nums[i] <= 10^5

    2228ms

     1 class Solution {
     2     func canDivideIntoSubsequences(_ nums: [Int], _ K: Int) -> Bool {
     3         var cal:Int = 0
     4         var a:[Int:Int] = [Int:Int]()
     5         for i in 0..<nums.count
     6         {
     7             a[nums[i],default:0] += 1
     8             if cal < a[nums[i],default:0]
     9             {
    10                 cal = a[nums[i],default:0]
    11             }
    12         }
    13         return cal * K <= nums.count
    14     }
    15 }
  • 相关阅读:
    FS配置文件
    FS 目录结构
    FreeSWITCH 架构
    FreeSWITCH API 与APP
    呼入电话处理
    FS 分机呼出
    PartyUs 待添加内容(不定期更新)
    PHP-Codeigniter:实习笔记5
    PHP-Codeigniter:实习笔记4
    PHP-Codeigniter:实习笔记3
  • 原文地址:https://www.cnblogs.com/strengthen/p/11179565.html
Copyright © 2020-2023  润新知