CodeWars 关于字符串包含问题

最新推荐文章于 2021-03-29 09:31:07 发布

原创最新推荐文章于 2021-03-29 09:31:07 发布 · 1k 阅读

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本文介绍了一种检查一个字符串是否可以通过重组其字符来匹配另一个字符串的方法。提供了两种不同的Java实现方案，一种通过逐字符比较并移除的方式，另一种使用了哈希映射进行计数对比。

Description:

Write function scramble(str1,str2) that returns true if a portion of str1 characters can be rearranged to match str2, otherwise returns false.

For example:
str1 is ‘rkqodlw’ and str2 is ‘world’ the output should return true.
str1 is ‘cedewaraaossoqqyt’ and str2 is ‘codewars’ should return true.
str1 is ‘katas’ and str2 is ‘steak’ should return false.

Only lower case letters will be used (a-z). No punctuation or digits will be included.
Performance needs to be considered

public class Scramblies {

    public static boolean scramble(String str1, String str2) {
        if (str2.length() > str1.length()) return false;
        for (String s: str2.split("")) {
          if (!str1.contains(s))  return false;
          str1 = str1.replaceFirst(s,"");
        }        

        return true;
    }
}

另一种实现方式

import java.util.HashMap;
import java.util.Map;

public class Scramblies {

    public static boolean scramble(String str1, String str2) {
        if (str1.length() < str2.length()) return false;
        Map requirements = new HashMap<>();
        for (char reqChar : str2.toCharArray()) {
            Integer count = requirements.get(reqChar);
            if (count == null) count = 0;
            requirements.put(reqChar, count + 1);
        }

        for (char foundChar : str1.toCharArray()) {
            Integer count = requirements.remove(foundChar);
            if (count == null) continue;
            if (count > 1) requirements.put(foundChar, count - 1);
        }
        return requirements.isEmpty();
    }
}