Java数组元素搜索：如何正确判断元素不存在（含代码问题）

需求：实现Java程序，向用户索要待搜索数字，若数组包含该数字则输出其索引；若不包含则提示数字未找到。

现有代码能判断元素存在，但判断不存在的逻辑存在错误：

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int[] array = new int[8];
        array[0] = 6;
        array[1] = 2;
        array[2] = 8;
        array[3] = 1;
        array[4] = 3;
        array[5] = 0;
        array[6] = 9;
        array[7] = 7;

        System.out.print("Search for? ");
        int searchValueInput = scanner.nextInt();

        for (int loopValue = 0; loopValue < array.length; loopValue++) {
            if (searchValueInput == array[loopValue]) {
                System.out.println(searchValueInput + " is at index " + loopValue + ".");
            } else if (searchValueInput > array.length) {
                System.out.println(searchValueInput + " was not found.");
            }
        }
    }
}

错误分析 #

1. 判断searchValueInput > array.length完全不合理：数组长度是8，但数组元素可以是任意整数，比如元素9大于8却存在于数组中，22大于8但不存在，这种逻辑完全不成立。
2. 循环中每次不匹配都会触发else if，导致找不到元素时会重复输出多次"未找到"提示。

修复方案 #

用布尔变量标记是否找到目标元素，遍历完数组后再根据标记判断是否输出"未找到"；另外找到元素后直接跳出循环，避免不必要的遍历：

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int[] array = {6, 2, 8, 1, 3, 0, 9, 7}; // 简化数组初始化

        System.out.print("Search for? ");
        int searchValueInput = scanner.nextInt();
        
        boolean found = false;

        for (int loopValue = 0; loopValue < array.length; loopValue++) {
            if (searchValueInput == array[loopValue]) {
                System.out.println(searchValueInput + " is at index " + loopValue + ".");
                found = true;
                break; // 找到后直接终止循环
            }
        }

        // 遍历结束后统一判断是否未找到
        if (!found) {
            System.out.println(searchValueInput + " was not found.");
        }
    }
}

期望输出示例 #

Search for? 3
3 is at index 4.

Search for? 9
9 is at index 6.

Search for? 22
22 was not found.

内容的提问来源于stack exchange，提问作者belinyom