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文字列「00010」を「10」にする方法の速さ比較

Last updated at Posted at 2016-09-02

文字列「00010」を「10」に変換する方法をいくつか考え、速さを比較しました。

前提条件は以下です。

  • 空文字の入力は想定しない。
  • 先頭の0を除く文字列はint型に収まる数値。
  • 以下のように先頭から余計な0を除く。
    • 00010 -> 10
    • 10010 -> 10010
    • 00 -> 0
    • 0 -> 0
    • 1 -> 1

以下の方法を比較しました。

  • Integer#parseInt()メソッドを使用
  • BigDecimalを使用
  • String#replaceAll()メソッドを使用

実行環境は以下の通りです。

  • DELL VOSTRO 1540
  • Windows 10 Pro 32bit
  • Intel Celelron 2.00 GHZ
  • メモリ 2.0GB
  • HDは約300GB

比較に使ったソースは以下です。

Java
public class HelloPrefixTrim {

    private static final String REG1 = "^0+";
    private static final String REP1 = "";

    private static final String REG2 = "^0+$";
    private static final String REP2 = "0";

    private static final String s1 = "00010";
    private static final String s2 = "10010";
    private static final String s3 = "00";
    private static final String s4 = "0";
    private static final String s5 = "1";

    private static int COUNT = 200000;

    public static void main(String[] args) {

        long start, end;

        start = System.currentTimeMillis();
        test1();
        end   = System.currentTimeMillis();
        System.out.println(end - start);

        start = System.currentTimeMillis();
        test2();
        end   = System.currentTimeMillis();
        System.out.println(end - start);

        start = System.currentTimeMillis();
        test3();
        end   = System.currentTimeMillis();
        System.out.println(end - start);

    }

    private static void test1() {

        for (int i = 0; i < COUNT; i++) {
            String.valueOf(Integer.parseInt(s1));
            String.valueOf(Integer.parseInt(s2));
            String.valueOf(Integer.parseInt(s3));
            String.valueOf(Integer.parseInt(s4));
            String.valueOf(Integer.parseInt(s5));
        }

    }

    private static void test2() {

        for (int i = 0; i < COUNT; i++) {
            new BigDecimal(s1).toString();
            new BigDecimal(s2).toString();
            new BigDecimal(s3).toString();
            new BigDecimal(s4).toString();
            new BigDecimal(s5).toString();
        }

    }

    private static void test3() {

        for (int i = 0; i < COUNT; i++) {
            replaceAll(s1);
            replaceAll(s2);
            replaceAll(s3);
            replaceAll(s4);
            replaceAll(s5);
        }

    }

    private static String replaceAll(String s) {
        if (s.matches(REG2)) {
            return REP2;
        } else {
            return s.replaceAll(REG1, REP1);
        }

    }
}
結果
119
164
1098

Integer#parseInt()メソッドが一番速いという結果になりました。

#@shiracamus さんによる処理との比較

コメントにて、@shiracamus さんに独自の処理を教えていただきました。以下、追加して結果を載せました。

HelloPrefixTrim2
public class HelloPrefixTrim2 {

    private static final String REG1 = "^0+";
    private static final String REP1 = "";

    private static final String REG2 = "^0+$";
    private static final String REP2 = "0";

    private static final String s1 = "00010";
    private static final String s2 = "10010";
    private static final String s3 = "00";
    private static final String s4 = "0";
    private static final String s5 = "1";

    private static int COUNT = 200000;

    public static void main(String[] args) {

        long start, end;

        start = System.currentTimeMillis();
        test1();
        end   = System.currentTimeMillis();
        System.out.println(end - start);

        start = System.currentTimeMillis();
        test2();
        end   = System.currentTimeMillis();
        System.out.println(end - start);

        start = System.currentTimeMillis();
        test3();
        end   = System.currentTimeMillis();
        System.out.println(end - start);

        start = System.currentTimeMillis();
        test4();
        end   = System.currentTimeMillis();
        System.out.println(end - start);
    }

    private static void test1() {

        for (int i = 0; i < COUNT; i++) {
            String.valueOf(Integer.parseInt(s1));
            String.valueOf(Integer.parseInt(s2));
            String.valueOf(Integer.parseInt(s3));
            String.valueOf(Integer.parseInt(s4));
            String.valueOf(Integer.parseInt(s5));
        }

    }

    private static void test2() {

        for (int i = 0; i < COUNT; i++) {
            new BigDecimal(s1).toString();
            new BigDecimal(s2).toString();
            new BigDecimal(s3).toString();
            new BigDecimal(s4).toString();
            new BigDecimal(s5).toString();
        }

    }

    private static void test3() {

        for (int i = 0; i < COUNT; i++) {
            replaceAll(s1);
            replaceAll(s2);
            replaceAll(s3);
            replaceAll(s4);
            replaceAll(s5);
        }

    }

    private static String replaceAll(String s) {
        if (s.matches(REG2)) {
            return REP2;
        } else {
            return s.replaceAll(REG1, REP1);
        }

    }

    private static void test4() {

        for (int i = 0; i < COUNT; i++) {
            trimzero(s1);
            trimzero(s2);
            trimzero(s3);
            trimzero(s4);
            trimzero(s5);
        }

    }

    private static String trimzero(String s) {
        int i = 0, last = s.length() - 1;
        while (i < last && s.charAt(i) == '0') i++;
        return s.substring(i);
    }
}
121
192
1121
30

@shiracamus さんの方法が圧倒的に速いです。

#修正履歴

20160902

  • 入力が空文字や0の場合を考慮し、全体的に修正。
  • @shiracamus さんに教えていただいたプログラムを追加。
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