如何对包含Java中的数字的字符串进行排序

Question

我想对具有nr的String进行排序.我怎么做？

让我们说我的整数是

Class2
"3"
"4"
"1"

在main我做class2.Sort();

提前致谢.

Answer 1

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

在字符串可能包含数字运行的情况下,自然排序实际上非常重要,并且您希望按字母顺序对字母进行排序,但在数字上按字母顺序排序.例如,现代版本的Windows资源管理器使用它来订购文件名.根据版本字符串选择最新版本的库(即"1.2.3"与"1.20.1"相比)也非常方便.

如果你的字符串真的只包含数字(就像你在描述中所说的那样),那么你最好不要使用字符串 - 而是创建并使用Integer对象.

注意:上面的链接似乎已被破坏.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 <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 2

  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 3

您的问题格式不正确，但您应该了解以下几点：

• 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[].

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。

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。

让我们假设这BigInteger是必需的。

您现在有两个选择：

• 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。