Java 不使用字符串函数的字符串中子字符串的出现次数
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Occurrences of a substring in a string without using string functions
提问by RTG87
I want to know how to count the occurrences of a particular sub-string in a string without using any of the built in JAVA string functions. For example:
我想知道如何在不使用任何内置 JAVA 字符串函数的情况下计算字符串中特定子字符串的出现次数。例如：
InputString = "knowbutuknow"
subString = "know"
The program should return the result as 2.
程序应该返回结果为 2。
EDIT: Re-phrased my question. This is one of those interview questions I came across.
编辑：重新表述我的问题。这是我遇到的面试问题之一。
EDIT: Basic string functions like charAt and length can be used here.
编辑：可以在此处使用基本字符串函数，如 charAt 和 length。
采纳答案by dognose
Assuming, you already know the keyword you are searching for:
假设您已经知道要搜索的关键字：
- Start at char "0" of Input String
- Iterate until "length - keyWordLength" (keyword of length 4 can not match into the last 3 chars)
- Inside: Iterate from 0 to keyWord.length -1 and always compare:
- Char at Position of outer loop PLUS position of inner loop of the Input string with the char at "inner loops" position of the keyword.
- if you find a match, go ahead with the innerloop, if it does not match, advance the outer loop, by simple "breaking" the inner loop.
- If you have a match, and completely processed the inner loop, you have a match of that keyword.
- 从输入字符串的字符“0”开始
- 迭代直到“length - keyWordLength”（长度为4的关键字不能匹配到最后3个字符）
- 内部：从 0 迭代到 keyWord.length -1 并始终比较：
- 外循环位置的字符加上输入字符串的内循环位置，字符位于关键字的“内循环”位置。
- 如果找到匹配项，则继续进行内循环，如果不匹配，则通过简单的“破坏”内循环来推进外循环。
- 如果您有匹配项，并且完全处理了内部循环，那么您就有了该关键字的匹配项。
Something like this. I'm Assuming String.lengthto be allowed. Otherwhise you would need to create your own strlen function. (This can be achieved, using a forachloop and simple counting "up")
像这样的东西。我假设String.length被允许。否则，您需要创建自己的 strlen 函数。（这可以实现，使用forach循环和简单的“向上”计数）
This is untested and may not work out of the box, but should give you a brief idea.
这是未经测试的，可能无法开箱即用，但应该给您一个简短的想法。
String inputString = "knowbutuknow";
String subString = "know";
int matches = 0;
for (int outer = 0; outer <= inputString.length() - subString.length(); outer++){
for (int inner = 0; inner < subString.length(); inner++){
if (inputString.charAt(outer + inner) == subString.charAt(inner)){
// letter matched, proceed.
if (inner == subString.length()-1){
//last letter matched, so a word match at position "outer"
matches++;
//proceed with outer. Room for improvement: Skip next n chars beeing
// part of the match already.
break;
}
}else{
//no match for "outer" position, proceed to next char.
break;
}
}
}
edit: Sorry, mixed in some php :) fixed it.
编辑：抱歉，混入了一些 php :) 修复了它。
