我想对具有nr的String进行排序。我该怎么做？
让我说我的整数是
Class2
"3"
"4"
"1"
在main中我做class2.Sort（）;
先谢谢。
答案 0 :(得分：17)
通用解决方案是使用所谓的“自然顺序比较器”。
以下是一个例子：
http://pierre-luc.paour.9online.fr/NaturalOrderComparator.java
自然排序实际上非常重要，因为字符串可能包含数字运行，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);
}
}
}
答案 1 :(得分：14)
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));
}
答案 2 :(得分：3)
你的问题相当不明确，但这里有一些你应该知道的事情：
Arrays.sort(Object[])对任意对象进行排序。String的自然顺序为lexicographic。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"），则只需Arrays.sort(sarr);即可。字符串是否包含数字无关紧要。
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字值。根据这些数字的范围，Integer.parseInt可能会这样做;你可以随时使用BigInteger。
我们假设BigInteger是必需的。
您现在有两个选择：
将String[]转换为BigInteger[]，然后从BigInteger implements Comparable<BigInteger>开始，您可以使用Arrays.sort使用其自然顺序。然后，您可以将已排序的BigInteger[]转换回String[]。
将String转换为BigInteger“及时”，以便通过自定义Comparator<String>进行比较。由于Arrays.sort使用基于比较的mergesort，因此您可以进行O(N log N)次比较，因此需要进行多次转换。
答案 3 :(得分：1)
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());
}
答案 4 :(得分：0)
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并对生成的数组进行排序。排序的函数是Arrays.sort（），如果内存为我服务
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
答案 5 :(得分：0)
根据某种顺序对“事物”进行排序的方法是创建一个比较器，它根据顺序知道哪一个是第一个，或者让“事物”本身实现Comparable接口，这样你就不需要了比较者。
如果你的工作是整数排序，那么考虑将转换为整数，然后然后排序，因为Integer类已经实现了Comparable。
答案 6 :(得分：0)
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）;