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java - 如何在 Java 中对包含数字的字符串进行排序

转载 作者:搜寻专家 更新时间:2023-11-01 03:57:20 31 4
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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"

主要我做 class2.Sort();

提前致谢。

最佳答案

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

在字符串可能包含连续数字并且您希望事物按字母顺序排序但按数字顺序排序的情况下,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)挑选最新版本的库也非常方便。

如果您的字符串真的只包含数字(就像您在描述中所做的那样),那么您最好根本不使用字符串 - 而是创建和使用 Integer 对象。

注意:上面的链接好像失效了。该代码非常有用,我将把它贴在这里:

/*
* <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);
}
}


}

关于java - 如何在 Java 中对包含数字的字符串进行排序,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/2529045/

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