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我如何对 Java的数字进行描述。
原标题:How do I sort strings that contain numbers in Java

我想打上一条 n子。 我如何这样做?

让我说一下我的愤怒。

Class2
"3"
"4"
"1"

a. 在主要一级,2.Sort();

预付款。

最佳回答
  public static void main(String[] args)
  {
    String string = "3 42 
   11   	  7  dsfss  365          
   1";
    String[] numbers = string.split("\D+");
    Arrays.sort(numbers, new Comparator<String>()
    {
      public int compare(String s1, String s2)
      {
        return Integer.valueOf(s1).compareTo(Integer.valueOf(s2));
      }
    });
    System.out.println(Arrays.toString(numbers));
  }
问题回答

一种通用的解决办法是使用所谓的自然秩序比较器。

例如:

在扼杀可能包含大量数字的情况下,自然秩序实际上非常重要,而你想把一些东西用在字母上,但数字上。 例如,Windows探测器的现代版本用于订购文档名称。 该图书馆还非常精彩地挑选了一个基于版本的座标(即“1.2.3”而不是“1.20.1”)的图书馆最新版本。

如果你的说明确实含有数字(如你描述的那样),那么你最好不要使用插手——而是制造和与印地物体合作。

注:上述联系似乎被打破。 守则非常有用,我可以在这里:

/*
 * <copyright>
 *
 *  Copyright 1997-2007 BBNT Solutions, LLC
 *  under sponsorship of the Defense Advanced Research Projects
 *  Agency (DARPA).
 *
 *  You can redistribute this software and/or modify it under the
 *  terms of the Cougaar Open Source License as published on the
 *  Cougaar Open Source Website (www.cougaar.org).
 *
 *  THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS
 *  "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT
 *  LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR
 *  A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT
 *  OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL,
 *  SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT
 *  LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE,
 *  DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY
 *  THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
 *  (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
 *  OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
 *
 * </copyright>
 */
/*
NaturalOrderComparator.java -- Perform  natural order  comparisons of strings in Java.
Copyright (C) 2003 by Pierre-Luc Paour <natorder@paour.com>

Based on the C version by Martin Pool, of which this is more or less a straight conversion.
Copyright (C) 2000 by Martin Pool <mbp@humbug.org.au>

This software is provided  as-is , without any express or implied
warranty.  In no event will the authors be held liable for any damages
arising from the use of this software.

Permission is granted to anyone to use this software for any purpose,
including commercial applications, and to alter it and redistribute it
freely, subject to the following restrictions:

1. The origin of this software must not be misrepresented; you must not
claim that you wrote the original software. If you use this software
in a product, an acknowledgment in the product documentation would be
appreciated but is not required.
2. Altered source versions must be plainly marked as such, and must not be
misrepresented as being the original software.
3. This notice may not be removed or altered from any source distribution.
 */
package org.cougaar.util;

//CHANGES: KD - added case sensitive ordering capability
// Made comparison so it doesn t treat spaces as special characters

//CHANGES:
//   set package to "org.cougaar.util"
//   replaced "import java.util.*" with explicit imports,
//   added "main" file reader support

import java.util.Comparator;

/**
 * A sorting comparator to sort strings numerically,
 * ie [1, 2, 10], as opposed to [1, 10, 2].
 */
public final class NaturalOrderComparator<T> implements  Comparator<T> {

    public static final Comparator<String> NUMERICAL_ORDER = new NaturalOrderComparator<String>(false);
    public static final Comparator<String> CASEINSENSITIVE_NUMERICAL_ORDER = new NaturalOrderComparator<String>(true);

    private final boolean caseInsensitive;

    private NaturalOrderComparator(boolean caseInsensitive) {
        this.caseInsensitive = caseInsensitive;
    }

    int compareRight(String a, String b) {
        int bias = 0;
        int ia = 0;
        int ib = 0;

        // The longest run of digits wins.  That aside, the greatest
        // value wins, but we can t know that it will until we ve scanned
        // both numbers to know that they have the same magnitude, so we
        // remember it in BIAS.
        for (;; ia++, ib++) {
            char ca = charAt(a, ia);
            char cb = charAt(b, ib);

            if (!Character.isDigit(ca) && !Character.isDigit(cb)) {
                return bias;
            } else if (!Character.isDigit(ca)) {
                return -1;
            } else if (!Character.isDigit(cb)) {
                return +1;
            } else if (ca < cb) {
                if (bias == 0) {
                    bias = -1;
                }
            } else if (ca > cb) {
                if (bias == 0)
                    bias = +1;
            } else if (ca == 0 && cb == 0) {
                return bias;
            }
        }
    }

    public int compare(T o1, T o2) {
        String a = o1.toString();
        String b = o2.toString();

        int ia = 0, ib = 0;
        int nza = 0, nzb = 0;
        char ca, cb;
        int result;

        while (true) {
            // only count the number of zeroes leading the last number compared
            nza = nzb = 0;

            ca = charAt(a, ia);
            cb = charAt(b, ib);

            // skip over leading zeros
            while (ca ==  0 ) {
                if (ca ==  0 ) {
                    nza++;
                } else {
                    // only count consecutive zeroes
                    nza = 0;
                }

                // if the next character isn t a digit, then we ve had a run of only zeros
                // we still need to treat this as a 0 for comparison purposes
                if (!Character.isDigit(charAt(a, ia+1)))
                    break;

                ca = charAt(a, ++ia);
            }

            while (cb ==  0 ) {
                if (cb ==  0 ) {
                    nzb++;
                } else {
                    // only count consecutive zeroes
                    nzb = 0;
                }

                // if the next character isn t a digit, then we ve had a run of only zeros
                // we still need to treat this as a 0 for comparison purposes
                if (!Character.isDigit(charAt(b, ib+1)))
                    break;

                cb = charAt(b, ++ib);
            }

            // process run of digits
            if (Character.isDigit(ca) && Character.isDigit(cb)) {
                if ((result = compareRight(a.substring(ia), b
                        .substring(ib))) != 0) {
                    return result;
                }
            }

            if (ca == 0 && cb == 0) {
                // The strings compare the same.  Perhaps the caller
                // will want to call strcmp to break the tie.
                return nza - nzb;
            }

            if (ca < cb) {
                return -1;
            } else if (ca > cb) {
                return +1;
            }

            ++ia;
            ++ib;
        }
    }

    private char charAt(String s, int i) {
        if (i >= s.length()) {
            return 0;
        } else {
            return caseInsensitive ? Character.toUpperCase(s.charAt(i)) : s.charAt(i);
        }
    }


}

你的问题很不了解,但这里是你们应当知道的几点:

因此,String[] sarr, 如果你想以图表方式分类(即1””和“10”和“10”及“2”,则简单明了Arrays.sort(sarr);。 如果指示数含有数量或没有数量,则确实如此。

如果你想将插图与编号(即1”和“2”及“2”和“10”)一样,那么你需要将插图转化为数字数值。 根据这些编号的范围,Integer.parse 可在上填写;否则,可以始终使用<条码>。

请允许我假设,需要<代码>BigInteger

你现在有两个选择:

  • Convert String[ ] to BigInteger[ ]/code>, 此后, BigInteger 执行 比较和比较;BigInteger>, 您可使用Arrays.sort, 采用其自然顺序。 之后,您可将打字的<代码>BigInteger[]重新改为String[

  • Convert String to BigInteger "just-in-time" for comparative by atom Comparator<String>。 自Arrays.sort 利用基于合并的比较,你可以预期(Nlog N)的比较,因此可以做许多换算。

Java8

public static List<String> sortAsNumbers(Collection<String> collection) {
    return collection
            .stream()
            .map(Integer::valueOf)
            .sorted()
            .map(String::valueOf)
            .collect(Collectors.toList());
}

根据某种命令分类“星号”的方法是建立一个比较器,该比较器知道任何两件事中哪一件事首先按照订单进行,或者让“星号”本身实施可比较的接口,因此不需要比较器。

如果您的工作是按分类的,则考虑将<>tverting改为 Integers和then,作为Integer类别已经实施的比较。

如果数字是所有一位数,则将座标分为代数阵列,并对阵列进行分类。 否则,必须有一个排位,以掌握数字。 呼唤。 使用该划界器进行分离,并对由此产生的阵列进行分类。 如果记忆为我服务的话,这种功能就是阿雷拉。

javadoc 参考

http://java.sun.com/j2se/1.4.2/docs/api/java/lang/String.html#split%28java.lang.String%29 http://java.sun.com/javase/6/docs/api/java/util/Arrays.html#sort%28double[]%29

static final Comparator<Object> COMPARADOR = new Comparator<Object>() {
    public int compare(Object o1, Object o2) {
        double numero1;
        double numero2;
        try {
            numero1 = Double.parseDouble(o1.toString());
            numero2 = Double.parseDouble(o2.toString());
            return Double.compare(numero1, numero2);
        } catch (Exception e) {
            return o1.toString().compareTo(o2.toString());
        }
    }
};

... ArrayList listaDeDatos; listaDeDatos.sort(COMPARADOR);





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