逐字符比较多个字符串java

Question

我想逐个字符地比较多个字符串，并希望输出一个字符串，其中包含大多数字符串中存在的字符。 例如我使用了三个字符串： 输入：S1=哪里，S2=这里，S3=哪里 输出：S=哪里。

我可以使用代码为 3 个字符串做到这一点：

public class stringc {

    static String S1="where";
    static String S2="wHere";
    static String S3="whera";
    static StringBuilder S=new StringBuilder();

    public static void main(String[] args)
    {
        for (int i=0;i<5;i++)
        {

            if(S1.charAt(i)==S2.charAt(i)||S1.charAt(i)==S3.charAt(i))
            {
                S.append(S1.charAt(i));
            }
        }
    }
}

谁能帮我用它来处理超过 10 个字符串。

Answer 1

使用数组/字符串列表，而不是 int i=0;i<5;i++ 使用 int i=0;i<array.length;i++ // 使用 list.size() 作为列表

Answer 2

这是我敲的一个。它比较输入字符串中的字符，以检查它们在相同位置的每个字符串中是否相同。 如果该位置的字符在所有输入字符串中都匹配，则相同的字符将出现在该位置的结果字符串中。

public class MultipleStringComparator {

public static void main(String[] args) {

    String[] input = new String[] {"abcdeFgh","abcdefgh","abcdefgh","abcdefgh","abcdefgh"};

    MultipleStringComparator comp = new MultipleStringComparator();
    System.out.println(comp.compare(input, '.'));
}

/**
 * Compare strings character by character
 * @param strings
 * @param substitute - character to insert in result where strings do not match
 * @return result
 */
public String compare(String[] strings, char substitute) {

    //How long is the longest string?
    int longest = 0;
    for (String string : strings) {
        if (string.length()>longest) longest = string.length();
    }

    //Initialise result
    StringBuffer result = new StringBuffer();

    //compare strings character by character
    //If the corresponding characters in all input strings match, append to result
    for (int position=0; position<longest; position++) {
        char character = allCharactersMatch(strings, position);
        if (character!=0) {
            result.append(character);
        } else {
            result.append(substitute);
        }
    }

    return result.toString();
}

/**
 * Compares the character at the specified position in all input strings
 * @param strings
 * @param position
 * @return character found is same in all strings, otherwise 0;
 */
private char allCharactersMatch(String[] strings, int position) {
    char found = 0;
    for (String string : strings) {
        if (string.length()<=position) return 0;
        if (string.charAt(position)!=found && found!=0) return 0;
        found = string.charAt(position);
    }
    return found;
}

}

Answer 3

我会怎么做：

1. 对于您的每个字符串，请致电toCharArray()
2. 将生成的 char[] 添加到 2D char 数组中
3. 对于 2D 数组中的每个单词，将您正在检查的字符添加到映射中，如果字符已经在映射 keySet() 中，则递增计数器
4. 地图中出现次数最多的字符（最高计数器）是要添加到结果字符串中的字符。
5. 对字符串中的所有字符重复步骤 3 和 5
6. 输出你构建的字符串

这是一段代码，使用您的示例和更多字符串：

public static void main(final String[] args) {
    final  StringBuilder stringBuilder = new StringBuilder();

    //Points 1 and 2
    final char[][] characters = new char[][]{
            "where".toCharArray(),
            "wHere".toCharArray(),
            "where".toCharArray(),
            "wperg".toCharArray(),
            "where".toCharArray(),
            "w6ere".toCharArray(),
            "where".toCharArray(),
            "where".toCharArray(),
            "wHere".toCharArray(),
            "w4eeg".toCharArray(),
            "where".toCharArray(),
            "wHare".toCharArray(),
            "where".toCharArray(),
            "where".toCharArray(),
            "weede".toCharArray(),
            "whare".toCharArray(),
            "wHect".toCharArray(),
            "where".toCharArray(),
            "wHere".toCharArray(),
            "whara".toCharArray()
    };

    //Point 3
    for (int i=0; i < 5 ; i++) {

        //Point 4
        final Map<String, Integer> occurrences = new HashMap<String, Integer>();
        for(int j = 0; j < characters.length; j++ ) {

            final String character = ""+characters[j][i];

            if( occurrences.containsKey(character) ) {
                Integer currentTotal = occurrences.get(character);
                occurrences.put(character, ++currentTotal);
                continue;
            }

            occurrences.put(character, 1);
        }

        //Point 5
        int mostOccurrences = 0;
        String characterWithMostOccurrences = "";
        for (final String character : occurrences.keySet()) {
            if( occurrences.get(character) > mostOccurrences ) {
                mostOccurrences = occurrences.get(character);
                characterWithMostOccurrences = character;
            }
        }

        stringBuilder.append(characterWithMostOccurrences);

    }

    //Point 6
    System.out.println(stringBuilder.toString());
}