zoukankan      html  css  js  c++  java
  • [Swift]LeetCode466. 统计重复个数 | Count The Repetitions

    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
    ➤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/10346034.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!
    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

    Define S = [s,n] as the string S which consists of n connected strings s. For example, ["abc", 3]="abcabcabc".

    On the other hand, we define that string s1 can be obtained from string s2 if we can remove some characters from s2 such that it becomes s1. For example, “abc” can be obtained from “abdbec” based on our definition, but it can not be obtained from “acbbe”.

    You are given two non-empty strings s1 and s2 (each at most 100 characters long) and two integers 0 ≤ n1 ≤ 106 and 1 ≤ n2 ≤ 106. Now consider the strings S1 and S2, where S1=[s1,n1] and S2=[s2,n2]. Find the maximum integer M such that [S2,M] can be obtained from S1.

    Example:

    Input:
    s1="acb", n1=4
    s2="ab", n2=2
    
    Return:
    2

    定义由 n 个连接的字符串 s 组成字符串 S,即 S = [s,n]。例如,["abc", 3]=“abcabcabc”。

    另一方面,如果我们可以从 s2 中删除某些字符使其变为 s1,我们称字符串 s1 可以从字符串 s2 获得。例如,“abc” 可以根据我们的定义从 “abdbec” 获得,但不能从 “acbbe” 获得。

    现在给出两个非空字符串 S1 和 S2(每个最多 100 个字符长)和两个整数 0 ≤ N1 ≤ 106 和 1 ≤ N2 ≤ 106。现在考虑字符串 S1 和 S2,其中S1=[s1,n1]S2=[s2,n2]。找出可以使[S2,M]从 S1 获得的最大整数 M。

    示例:

    输入:
    s1 ="acb",n1 = 4
    s2 ="ab",n2 = 2
    
    返回:
    2

    132ms
     1 class Solution {
     2     func getMaxRepetitions(_ s1: String, _ n1: Int, _ s2: String, _ n2: Int) -> Int {
     3         var repeatCnt:[Int] = [Int](repeating:0,count:n1 + 1)
     4         var nextIdx:[Int] = [Int](repeating:0,count:n1 + 1)
     5         var j:Int = 0
     6         var cnt:Int = 0
     7         if n1 == 0 {return 0}
     8         for k in 1...n1
     9         {
    10             for i in 0..<s1.count
    11             {
    12                 if s1[i] == s2[j]
    13                 {
    14                     j += 1
    15                     if j == s2.count
    16                     {
    17                         j = 0
    18                         cnt += 1
    19                     }
    20                 }
    21             }
    22             repeatCnt[k] = cnt
    23             nextIdx[k] = j
    24             for start in 0..<k
    25             {
    26                 if nextIdx[start] == j
    27                 {
    28                     var interval:Int = k - start
    29                     var rep:Int = (n1 - start) / interval
    30                     var patternCnt:Int = (repeatCnt[k] - repeatCnt[start]) * rep
    31                     var remainCnt:Int = repeatCnt[start + (n1 - start) % interval]
    32                     return (patternCnt + remainCnt) / n2
    33                 }
    34             }
    35         }
    36         return repeatCnt[n1] / n2
    37     }
    38     
    39     func getArray(_ str:String) -> [Character]
    40     {
    41         var arr:[Character] = [Character]()
    42         for char in str.characters
    43         {
    44             arr.append(char)
    45         }
    46         return arr
    47     }
    48 }
    49 
    50 extension String {        
    51     //subscript函数可以检索数组中的值
    52     //直接按照索引方式截取指定索引的字符
    53     subscript (_ i: Int) -> Character {
    54         //读取字符
    55         get {return self[index(startIndex, offsetBy: i)]}
    56     }
    57 }
  • 相关阅读:
    让PHP程序永远在后台运行
    discuz3.2x增加邮箱验证功能
    UML类图几种关系的总结
    UML中九种图的理解
    什么是UML类图
    仿了么项目,商品详情页开发
    仿饿了么项目,右侧商品组件动画,以及和购物车组件的联动效果,小球掉落效果
    外卖项目底部购物车组件编写
    仿饿了么外卖项目better-scroll插件的实战
    vue项目如何在手机上测试
  • 原文地址:https://www.cnblogs.com/strengthen/p/10346034.html
Copyright © 2011-2022 走看看