如何在Java中对包含数字的字符串进行排序

Question

我想对具有 nr 的字符串进行排序。我该怎么做？

假设我的整数是

Class2
"3"
"4"
"1"

我主要做 class2.Sort();

提前致谢。

Answer 1

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);

Answer 2

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());
}

Answer 3

通用解决方案是使用所谓的“自然顺序比较器”。

这是一个例子：

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

在字符串可能包含一系列数字并且您希望按字母顺序对字母排序但在数字上按数字排序的情况下，自然排序实际上非常重要。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）选择最新版本的库也非常方便。

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。

注意：上面的链接似乎已损坏。该代码非常有用，我将在此处发布：

/*
 * <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 <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);
        }
    }


}

Answer 4

按照某种顺序对“事物”进行排序的方法是创建一个比较器，它可以根据顺序知道任意两个事物中的哪一个，或者让“事物”本身实现 Comparable 接口，因此您不需要比较器。

如果您的工作是按整数排序，请考虑将转换为整数并然后排序，因为 Integer 类已经实现了 Comparable。

Answer 5

  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));
  }

Answer 6

您的问题格式不正确，但您应该知道以下几点：

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

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

如果要将字符串排序为数字（即"1" < "2" < "10"），则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；否则，您始终可以使用BigInteger。

假设BigInteger 是必需的。

您现在有两个选择：

• 将String[] 转换为BigInteger[]，然后由于BigInteger implements Comparable<BigInteger>，您可以使用Arrays.sort 使用其自然顺序。然后您可以将排序后的BigInteger[] 转换回String[]。

• 将String 转换为BigInteger“即时”，以便通过自定义Comparator<String> 进行比较。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，因此会有尽可能多的转化。

Answer 7

如果数字都是单个数字，则将字符串拆分为 char 数组并对数组进行排序。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。如果我没记错的话，排序函数是 Arrays.sort()

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