在核心 Java 中使用递归机制在另一个字符串中查找字符串 [英] Finding a String in another String Using Recursion Mechanism in Core Java

问题描述

我有以下问题陈述

PS:给定一个字符串str"和一个非空子字符串sub"，如果sub"的至少N"个副本出现在字符串某处"中，则计算递归"，可能带有重叠".N 将为非负.

示例如下
strCopies("catcowcat", "cat", 2) → true
strCopies("catcowcat", "cow", 2) → false
strCopies("catcowcat", "cow", 1) → true
strCopies("iiijjj", "ii", 2) → true

我已经编写了如下所示的代码(没有递归)并且对于少数测试用例运行良好，除了其他标记为 FAIL 的测试用例.

I have written the code as shown below(without recursion) and is working fine for few test cases,except for others which are marked as FAIL.

:::代码如下图:::

public boolean strCopies(String str, String sub, int n) {    
    int len = sub.length();    
    int result=0;    
    if(len>0){    
       int start = str.indexOf(sub);    
       while(start !=-1){    
              result++;    
              start = str.indexOf(sub,start+len);                     
       }
    }          
   if(result==n){
        return true;
   }else return false; 
}

运行上述代码如下(粗体标记为失败的测试用例)

Runs for above code as shown below(Marked in BOLD are FAILED TEST CASES)

预期本次运行
strCopies("catcowcat", "cat", 2) → true true OK
strCopies("catcowcat", "cow", 2) → false false OK
strCopies("catcowcat", "cow", 1) → true true OK
strCopies("iiijjj", "ii", 2) → 真假失败
strCopies("iiiiij", "iii", 3) → 真假失败
strCopies("ijiiiiij", "iiii", 2) → 真假失败

你能检查一下并让我知道 FAIL TEST CASES 的代码有什么问题吗?我无法考虑重叠场景.

推荐答案

嗯，你的第一个问题是你的方法不是递归的.我怀疑您想使用 substring 以及 indexOf...

Well, your first problem is that your method isn't recursive. I suspect you want to work with substring as well as indexOf...

至于为什么您当前的方法不起作用，我怀疑是因为您使用 start + len 而不是 start + 1 来查找下一个起始位置.因此，当试图在iii"中找到ii"时，您应该首先查看位置 0，然后查看位置 1——目前您正在查看位置 2，这意味着它不会找到从 1 开始的第二个ii".

As for why your current method isn't working, I suspect it's because you're using start + len instead of start + 1 to find the next starting position. So when trying to find "ii" in "iii", you should first look at position 0, then position 1 - currently you're looking at position 2, which would mean it wouldn't find the second "ii" starting at 1.

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