Question

如何找到ArrayList中BigIntegers的最小和最大元素。

我尝试过的是：

import java.math.BigInteger; 
import java.io.*;
import java.util.*;

public class HelloWorld{

     public static void main(String []args){
         BigInteger i1 = new BigInteger("4343345345345");
         BigInteger i2 = new BigInteger("4343453345345345");
         BigInteger i3 = new BigInteger("4343453345");

        List<BigInteger> list = new ArrayList<>();

        list.add(i1);
        list.add(i2);
        list.add(i3);

       BigInteger max = list.stream().max(Comparator.comparing(BigInteger::intValue())).orElseThrow(NoSuchElementExcep::new);        

        BigInteger min = list.stream().min(Comparator.comparing(BigInteger::intValue())).orElseThrow(NoSuchElementExcep::new);

      System.out.println(max.intValue());
      System.out.println(min.intValue());
     }
}

但这给了我以下错误：

HelloWorld.java:20: error: ')' expected
        BigInteger min = list.stream().min(Comparator.comparing(BigInteger::intValue())).orElseThrow(NoSuchElementExcep::new);
                                                                                    ^
HelloWorld.java:20: error: ';' expected
        BigInteger min = list.stream().min(Comparator.comparing(BigInteger::intValue())).orElseThrow(NoSuchElementExcep::new);
                                                                                     ^
HelloWorld.java:20: error: illegal start of expression
        BigInteger min = list.stream().min(Comparator.comparing(BigInteger::intValue())).orElseThrow(NoSuchElementExcep::new);
                                                                                      ^
HelloWorld.java:20: error: ';' expected
        BigInteger min = list.stream().min(Comparator.comparing(BigInteger::intValue())).orElseThrow(NoSuchElementExcep::new);
                                                                                       ^
HelloWorld.java:20: error: illegal start of expression
        BigInteger min = list.stream().min(Comparator.comparing(BigInteger::intValue())).orElseThrow(NoSuchElementExcep::new);
                                                                                        ^
HelloWorld.java:20: error: ';' expected
        BigInteger min = list.stream().min(Comparator.comparing(BigInteger::intValue())).orElseThrow(NoSuchElementExcep::new);
                                                                                                    ^
HelloWorld.java:20: error: not a statement
        BigInteger min = list.stream().min(Comparator.comparing(BigInteger::intValue())).orElseThrow(NoSuchElementExcep::new);
                                                                                                     ^
HelloWorld.java:20: error: ';' expected
        BigInteger min = list.stream().min(Comparator.comparing(BigInteger::intValue())).orElseThrow(NoSuchElementExcep::new);

还有其他想法吗？

Answer 1

在不比较intValue的情况下获取最小值和最大值是可行的。您应该使用自然顺序：

BigInteger max = list.stream().max(BigInteger::compareTo).orElseThrow(NoSuchElementExcep::new);        
BigInteger min = list.stream().min(BigInteger::compareTo).orElseThrow(NoSuchElementExcep::new);

这也不是方法，您正在打印值：

System.out.println(max.intValue());
System.out.println(min.intValue());

您正在调用intValue方法，但是该数字超出了int（32b）的容量。您应该使用：

System.out.println(max);
System.out.println(min);

Answer 2

简单地做

# The search algorithm is actually only applied a filter on the objects array.

# data is all items in database

data = realm.objects(CustomObject.self).filter(filterPredicate(parentID: id, keyword: keyword, colorIndex: colorIndexes, isActionOnly: True)).sorted(by: descSorting)


private func filterPredicate(parentID: Int?, keyword: String?, colorIndex: [Int]?, isActionOnly: Bool = false) -> InspirationFilterClosure {
    return { item in
        if item.isDeleted { return false }
        if let id = parentID, item.parentID != id { return false }
        else if item.parentID < 0 { return false }
        if isActionOnly, !item.isAction  { return false }
        if let indexes = colorIndex, !indexes.contains(-1), !indexes.contains(item.colorIndex) { return false }
        if let keyword = keyword, !keyword.isEmpty {
            return item.content.range(of: keyword, options: .caseInsensitive) != nil
        }
        return true
    }
}

Answer 3

答案是：

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Answer 4

BigInteger max = list.stream().max(Comparator.comparing(val -> new BigInteger(String.valueOf(val))))
            .orElseThrow(NoSuchElementException::new);
BigInteger min = list.stream().min(Comparator.comparing(val -> new BigInteger(String.valueOf(val))))
            .orElseThrow(NoSuchElementException::new);

如何找到大整数数组列表的min和max元素？

4 个答案: