检查数组中是否存在元素 [英] Checking whether an element exist in an array

问题描述

用于检查数组中是否存在元素的简单Java代码：

A simple Java code for checking whether an element exists in an array or not:

import java.util.Arrays;

public class Main {
    static int[] numbers = {813, 907, 908, 909, 910};

    public static void main(String[] args) {
        int number = 907;
        //Integer number = 907; // the same thing -- it's not found.
        boolean b = Arrays.asList(numbers).contains(number);
        System.out.println(b);  // => false
    }
}

1）为什么不找到907 in数组？

1) Why doesn't it find 907 in the array?

2）如果有更好的方法，请继续分享您的知识。

2) If there is a better way of doing it, go ahead and share your knowledge.

更新：

据说 asList 转换你的 int [] 进入列表< int []> ，只有一个成员：原始列表。但是，我希望下面的代码能给我1，但它给了我5：

It was said that asList converts your int[] into a List<int[]> with a single member: the original list. However, I expect the following code to give me 1, but it gives me 5:

System.out.println(Arrays.asList(numbers).size());

推荐答案

问题在于数组.asList（数字）没有按照你的想法做。它将您的 int [] 转换为列表< int []> ，其中包含一个成员：原始列表。

The problem is that Arrays.asList(numbers) isn't doing what you think. It is converting your int[] into a List<int[]> with a single member: the original list.

您可以进行简单的线性搜索，或者如果您的数字数组始终排序，请使用 Arrays.binarySearch（numbers，907）; 并测试结果是否为负（意味着未找到）。

You can do a simple linear search or, if your numbers array is always sorted, use Arrays.binarySearch(numbers, 907); and test whether the result is negative (meaning not found).

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