如何在 Java 中对包含数字的字符串进行排序

How do I sort strings that contain numbers in Java

提问by Sanju

I want to sort a String that has nr. How do I do that?

我想对一个有 nr 的字符串进行排序。我怎么做？

Lets say my integers are

假设我的整数是

Class2
"3"
"4"
"1"

in main I do class2.Sort();

在主要我做 class2.Sort();

Thanks in Advance.

提前致谢。

采纳答案by eljenso

  public static void main(String[] args)
  {
    String string = "3 42 \n   11   \t  7  dsfss  365          \r   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));
  }

回答by Midhat

If the numbers are all single digit, split the string into a char array and sort the array. otherwise there has to be a delimiter to seperate the numbers. call string.split using that delimiter and sort the resulting array. The function to sort is Arrays.sort() if memory serves me right

如果数字都是一位数，则将字符串拆分为一个字符数组并对该数组进行排序。否则必须有一个分隔符来分隔数字。使用该分隔符调用 string.split 并对结果数组进行排序。排序的函数是 Arrays.sort() 如果我没记错的话

javadoc references

javadoc 引用

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

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

回答by polygenelubricants

Your question is rather ill-formed, but here are a few things that you should know:

您的问题格式不正确，但您应该了解以下几点：

• There is an Arrays.sort(Object[])that you can use to sort arbitrary objects.
• The natural ordering for Stringis lexicographic.
• Java arrays are covariant: a String[]is an Object[].

• 有一个Arrays.sort(Object[])可用于对任意对象进行排序。
• 的自然顺序String是lexicographic。
• Java 数组是协变的：aString[]是一个Object[].

So, given String[] sarr, if you want to sort it lexicographically (i.e. "1" < "10" < "2"), simply Arrays.sort(sarr);works. It doesn't matter if the strings contain numbers or not.

因此，给定String[] sarr，如果您想按字典顺序（即"1" < "10" < "2"）对其进行排序，则很简单Arrays.sort(sarr);。字符串是否包含数字并不重要。

If you want to sort the strings as if they are numbers (i.e. "1" < "2" < "10"), then you need to convert the strings to numeric values. Depending on the range of these numbers, Integer.parseIntmight do; you can always use BigIntegerotherwise.

如果您想对字符串进行排序，就好像它们是数字一样（即"1" < "2" < "10"），那么您需要将字符串转换为数值。根据这些数字的范围，Integer.parseInt可能会这样做；你总是可以使用BigInteger其他方式。

Let's assume that BigIntegeris required.

让我们假设这BigInteger是必需的。

You now have two options:

您现在有两个选择：

• Convert String[]to BigInteger[], then since BigInteger implements Comparable<BigInteger>, you can use Arrays.sortusing its natural ordering. You may then convert the sorted BigInteger[]back to String[].

• Convert Stringto BigInteger"just-in-time" for comparison by a custom Comparator<String>. Since Arrays.sortuses the comparison based mergesort, you can expect O(N log N)comparisons, and therefore as many conversions.

• 转换String[]为BigInteger[]，然后因为BigInteger implements Comparable<BigInteger>，您可以使用Arrays.sort其自然顺序。然后，您可以将排序BigInteger[]后的String[].

• 转换String为BigInteger“即时”以供自定义Comparator<String>. 由于Arrays.sort使用基于比较的归并排序，您可以期待O(N log N)比较，因此可以进行尽可能多的转换。

回答by Kevin Day

A general purpose solution is to use what's called a 'natural order comparator'.

通用解决方案是使用所谓的“自然顺序比较器”。

Here's an example:

下面是一个例子：

http://pierre-luc.paour.9online.fr/NaturalOrderComparator.java

http://pierre-luc.paour.9online.fr/NaturalOrderComparator.java

Natural ordering is actually quite important in cases where a string might contain runs of numbers and you want things to sort alphabetically on the letters but numerically on the numbers. Modern versions of Windows Explorer uses this for ordering file names, for example. It's also very handy for picking out the latest version of a library based on version strings (i.e. "1.2.3" compared to to "1.20.1").

在字符串可能包含一系列数字的情况下，自然排序实际上非常重要，并且您希望事物按字母顺序按字母排序，但按数字排序。例如，现代版本的 Windows 资源管理器使用它来排序文件名。根据版本字符串（即“1.2.3”与“1.20.1”相比）挑选库的最新版本也非常方便。

If your strings really just contain numbers (like you put in your description), then you are best off not using strings at all - create and work with Integer objects instead.

如果你的字符串真的只包含数字（就像你在描述中输入的那样），那么你最好不要使用字符串——而是创建和使用 Integer 对象。

Note: The link above seems to be broken. The code is so useful that I'm going to post it here:

注意：上面的链接似乎已损坏。代码非常有用，我将在这里发布：

/*
 * <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 <[email protected]>

Based on the C version by Martin Pool, of which this is more or less a straight conversion.
Copyright (C) 2000 by Martin Pool <[email protected]>

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);
        }
    }


}

回答by Thorbj?rn Ravn Andersen

The way to sort "things" according to some order is to create a Comparator which knows which of any two things are first according to the order, or to let the "thing" itself implement the Comparable interface so you do not need the Comparator.

根据某种顺序对“事物”进行排序的方法是创建一个 Comparator，它知道任何两个事物中的哪一个是按照顺序排在最前面的，或者让“事物”本身实现 Comparable 接口，这样您就不需要 Comparator。

If your job is to sort as integers, then consider convertingto integers and thensort as the Integer class already implements Comparable.

如果您的工作是按整数排序，则考虑转换为整数，然后按 Integer 类已经实现 Comparable进行排序。

回答by lobsten

In Java 8 we have nice solution

在 Java 8 中我们有很好的解决方案

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

回答by Hec T

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);

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