Most frequent k chars distance: Difference between revisions
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=={{header|Java}}== |
=={{header|Java}}== |
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{{incomplete|Java| Works now when there more than 10 occurences of a character in string. Please check, there can be more elegant way to deal with it. Comment was : This will fail catastrophically on “<tt>ABACADAEAFAEADACABA</tt>” as that has 10 ‘<tt>A</tt>’ characters in it.}} |
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Translation of the pseudo-code of the Wikipedia article [[wp:Most frequent k characters]] to [[wp:java]] implementation of three functions given in the definition section are given below with [[wp:JavaDoc]] comments: |
Translation of the pseudo-code of the Wikipedia article [[wp:Most frequent k characters]] to [[wp:java]] implementation of three functions given in the definition section are given below with [[wp:JavaDoc]] comments: |
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public static int getDiff(String str1, String str2, int limit) { |
public static int getDiff(String str1, String str2, int limit) { |
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int similarity = 0; |
int similarity = 0; |
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int k = 0; |
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for (int i = 0; i < str1.length() ; i = k) { |
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System.out.println(i); |
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if (Character.isLetter(str1.charAt(i))) { |
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similarity += Integer.parseInt(str2.substring(pos+1, pos+2)) + Character.getNumericValue(str1.charAt(i+1)); |
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} |
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String digitStr1 = ""; |
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while ( k < str1.length() && !Character.isLetter(str1.charAt(k))) { |
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digitStr1 += str1.charAt(k); |
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k++; |
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} |
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int k2 = pos+1; |
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String digitStr2 = ""; |
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while (k2 < str2.length() && !Character.isLetter(str2.charAt(k2)) ) { |
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digitStr2 += str2.charAt(k2); |
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k2++; |
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} |
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similarity += Integer.parseInt(digitStr2) |
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+ Integer.parseInt(digitStr1); |
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} |
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} |
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} |
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} |
} |
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/** |
/** |
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