Question

我正在处理一个向用户询问值的程序，并将它们添加到数组列表中，直到插入值为-1。然后它询问，您要搜索什么值，用户插入另一个值。然后程序在数组列表中搜索所有这些值，并打印出所有找到的项的索引。我的程序目前仅打印第一项及其索引，因为我使用list.indexOf（searchingitem）。如何让它列出所有找到的项目而不仅仅是第一项？到目前为止，这是我的代码。

import java.util.ArrayList;
import java.util.Scanner;

public class KysytynLuvunIndeksi {

    public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);

        ArrayList<Integer> list = new ArrayList<>();
        while (true) {
            int read = Integer.parseInt(reader.nextLine());
            if (read == -1) {
                break;
            }

            list.add(read);
        }

        System.out.print("What are you looking for? ");
        int searched = Integer.parseInt(reader.nextLine());

        if (list.contains(searched)) {
            System.out.println("Value " + searched + " has index of: " + (list.indexOf(searched));
        } else {
            System.out.println("value " + searched + " is not found");
        }

    }
}

提前致谢！

Answer 1

如果条件

，请将此循环放入

if (list.contains(searched)) {
     for (int index = list.indexOf(searched);index >= 0;index =list.indexOf(searched, index + 1))
    {
      System.out.println("Value " + searched + " has index of: " + index);
    }
}

Answer 2

遍历列表，将项目与searched进行比较，如果匹配则将索引添加到结果列表中。一些事情：

List<Integer> foundIndices = new ArrayList<>();
for (int index = 0; index < list.size(); index++) {
    if (list.get(index).intValue() == searched) {
        foundIndices.add(index);
    }
}

Answer 3

public static void main(String[] args) {
        Scanner reader = new Scanner(System.in);

        ArrayList<Integer> list = new ArrayList<>();
        while (true) {
            int read = Integer.parseInt(reader.nextLine());
            if (read == -1) {
                break;
            }

            list.add(read);
        }

        System.out.print("What are you looking for? ");
        int searched = Integer.parseInt(reader.nextLine());

        for(int i=0;i<list.size();i++){
            if(list.get(i)==searched){
                System.out.println("Value " + searched + " has index of: " + (i));
            }
        }
        if (!list.contains(searched)){
            System.out.println("value " + searched + " is not found");
       }    
    }

简单的for循环就足够了。

list.contains将返回数组列表中第一个找到的元素。

Answer 4

简单的for循环就足够了，并且比contains和indexOf的组合更快。您可以将循环索引保留在for语句之外，以检查之后是否找不到任何内容。

int i;
for (i = 0; i < list.size(); i++) {
    if (searched == list.get(i)) {
        System.out.println("Value " + searched + " has index of: " + i);
    }
}
if (i == list.size()) {
    System.out.println("value " + searched + " is not found");
}

Answer 5

如果你想使用.contains（）方法，我会建议这样做：

ArrayList<Integer> bufflist = list.clone(); // you create a temporal back of the list to operate with the same list without destroying the data
int searched = Integer.parseInt(reader.nextLine());

while (bufflist.contains(searched)) {
    System.out.println("Value " + searched + " has index of: " + (list.indexOf(searched));
     listbuff.remove(list.indexOf(searched)); //you remove from the list so it doesn´t find it again

}

Answer 6

尝试以下代码

GlobalKey

}

arrayList中项的Java索引

6 个答案: