使用 Java 在原语数组中查找最大值/最小值
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原文地址: http://stackoverflow.com/questions/1484347/
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Finding the max/min value in an array of primitives using Java
提问by Nick Heiner
It's trivial to write a function to determine the min/max value in an array, such as:
编写一个函数来确定数组中的最小值/最大值很简单，例如：
/**
*
* @param chars
* @return the max value in the array of chars
*/
private static int maxValue(char[] chars) {
int max = chars[0];
for (int ktr = 0; ktr < chars.length; ktr++) {
if (chars[ktr] > max) {
max = chars[ktr];
}
}
return max;
}
but isn't this already done somewhere?
但这不是已经在某处完成了吗？
采纳答案by Michael Rutherfurd
Using Commons Lang (to convert) + Collections (to min/max)
使用 Commons Lang（转换）+ Collections（最小/最大）
import java.util.Arrays;
import java.util.Collections;
import org.apache.commons.lang.ArrayUtils;
public class MinMaxValue {
public static void main(String[] args) {
char[] a = {'3', '5', '1', '4', '2'};
List b = Arrays.asList(ArrayUtils.toObject(a));
System.out.println(Collections.min(b));
System.out.println(Collections.max(b));
}
}
Note that Arrays.asList()wraps the underlying array, so it should not be too memory intensive and it should not perform a copy on the elements of the array.
请注意，Arrays.asList()包装底层数组，因此它不应占用太多内存，也不应在数组元素上执行复制。
回答by Bart Kiers
Yes, it's done in the Collectionsclass. Note that you will need to convert your primitive char array to a Character[] manually.
是的，它是在Collections类中完成的。请注意，您需要手动将原始字符数组转换为 Character[]。
A short demo:
一个简短的演示：
import java.util.*;
public class Main {
public static Character[] convert(char[] chars) {
Character[] copy = new Character[chars.length];
for(int i = 0; i < copy.length; i++) {
copy[i] = Character.valueOf(chars[i]);
}
return copy;
}
public static void main(String[] args) {
char[] a = {'3', '5', '1', '4', '2'};
Character[] b = convert(a);
System.out.println(Collections.max(Arrays.asList(b)));
}
}
回答by Christoph
Here's a utility class providing min/maxmethods for primitive types: Primitives.java
这是一个提供min/max基本类型方法的实用程序类：Primitives.java
回答by Andrew McKinlay
The Google Guava libraryhas min and max methods in its Chars, Ints, Longs, etc. classes.
在谷歌番石榴库有最大值和最小值的方法在它的字符数，类型int，long等类。
So you can simply use:
所以你可以简单地使用：
Chars.min(myarray)
No conversions are required and presumably it's efficiently implemented.
不需要转换，并且可以有效地实施。
回答by mark2
import java.util.Random;
public class Main {
public static void main(String[] args) {
int a[] = new int [100];
Random rnd = new Random ();
for (int i = 0; i< a.length; i++) {
a[i] = rnd.nextInt(99-0)+0;
System.out.println(a[i]);
}
int max = 0;
for (int i = 0; i < a.length; i++) {
a[i] = max;
for (int j = i+1; j<a.length; j++) {
if (a[j] > max) {
max = a[j];
}
}
}
System.out.println("Max element: " + max);
}
}
回答by whoduexpect
Pass the array to a method that sorts it with Arrays.sort()so it only sorts the array the method is using then sets minto array[0]and maxto array[array.length-1].
将数组传递给对其进行排序的方法，Arrays.sort()以便它仅对方法正在使用的数组进行排序，然后将min设置为array[0]和max为array[array.length-1]。
回答by Lubna_Nsour
import java.util.Arrays;
public class apples {
public static void main(String[] args) {
int a[] = {2,5,3,7,8};
Arrays.sort(a);
int min =a[0];
System.out.println(min);
int max= a[a.length-1];
System.out.println(max);
}
}
回答by Ortomala Lokni
You can simply use the new Java 8 Streamsbut you have to work with int.
你可以简单地使用新的Java 8Stream小号，但你必须有工作int。
The streammethod of the utility class Arraysgives you an IntStreamon which you can use the minmethod. You can also do max, sum, average,...
stream实用程序类的方法Arrays为您提供了IntStream可以使用该min方法的方法。你也可以做max, sum, average, ...
The getAsIntmethod is used to get the value from the OptionalInt
该getAsInt方法用于从OptionalInt
import java.util.Arrays;
public class Test {
public static void main(String[] args){
int[] tab = {12, 1, 21, 8};
int min = Arrays.stream(tab).min().getAsInt();
int max = Arrays.stream(tab).max().getAsInt();
System.out.println("Min = " + min);
System.out.println("Max = " + max)
}
}
==UPDATE==
==更新==
If execution time is important and you want to go through the data only once you can use the summaryStatistics()method like this
如果执行时间很重要并且您只想浏览一次数据，则可以使用这样的summaryStatistics()方法
import java.util.Arrays;
import java.util.IntSummaryStatistics;
public class SOTest {
public static void main(String[] args){
int[] tab = {12, 1, 21, 8};
IntSummaryStatistics stat = Arrays.stream(tab).summaryStatistics();
int min = stat.getMin();
int max = stat.getMax();
System.out.println("Min = " + min);
System.out.println("Max = " + max);
}
}
This approach can give better performance than classical loop because the summaryStatisticsmethod is a reduction operationand it allows parallelization.
这种方法可以提供比经典循环更好的性能，因为该summaryStatistics方法是一种归约操作，并且允许并行化。
回答by Kim G.
The basic way to get the min/max value of an Array. If you need the unsorted array, you may create a copy or pass it to a method that returns the min or max. If not, sorted array is better since it performs faster in some cases.
获取数组最小值/最大值的基本方法。如果您需要未排序的数组，您可以创建一个副本或将其传递给返回最小值或最大值的方法。如果没有，排序数组更好，因为它在某些情况下执行得更快。
public class MinMaxValueOfArray {
public static void main(String[] args) {
int[] A = {2, 4, 3, 5, 5};
Arrays.sort(A);
int min = A[0];
int max = A[A.length -1];
System.out.println("Min Value = " + min);
System.out.println("Max Value = " + max);
}
}
回答by winklerrr
You could easily do it with an IntStreamand the max()method.
您可以使用 anIntStream和max()方法轻松完成。
Example
例子
public static int maxValue(final int[] intArray) {
return IntStream.range(0, intArray.length).map(i -> intArray[i]).max().getAsInt();
}
Explanation
解释
range(0, intArray.length)- To get a stream with as many elements as present in theintArray.map(i -> intArray[i])- Map every element of the stream to an actual element of theintArray.max()- Get the maximum element of this stream asOptionalInt.getAsInt()- Unwrap theOptionalInt. (You could also use here:orElse(0), just in case theOptionalIntis empty.)
range(0, intArray.length)- 获取包含与intArray.map(i -> intArray[i])- 将流的每个元素映射到intArray.max()- 获取此流的最大元素为OptionalInt。getAsInt()- 解开OptionalInt. （您也可以在这里使用:orElse(0)，以防万一OptionalInt为空。）
