java - 排序数组中的线性搜索 - Java

Question

我想制作一个程序，在排序数组中进行线性搜索，并可以输出找到搜索项的不同位置。目前，我的程序仅输出找到搜索项的第一个位置，因此这里是我的程序现在所做的示例:

Enter number of elements
5
Enter 5 integers
1
3
3
9
15
Enter value to find
3
3 is present at location 2.

现在的问题是 3 位于位置 2 和 3 上，这就是我想在程序中编辑的内容，但我不知道该怎么做。

这是我的程序的代码:

import java.util.Scanner;
 class LinearSearchArray1 {
    public static void main(String args[]){
        int c, n, search, array[];

        Scanner in = new Scanner(System.in);
        System.out.println("Enter number of elements");
        n = in.nextInt(); 
        array = new int[n];

        System.out.println("Enter " + n + " integers");

        for (c = 0; c < n; c++)
        array[c] = in.nextInt();

        System.out.println("Enter value to find");
        search = in.nextInt();

        for (c = 0; c < n; c++)
        {
            if (array[c] == search)     /* Searching element is present */
            {
             System.out.println(search + " is present at location " + (c + 1) + ".");
            break;
        }
    }
    if (c == n)  /* Searching element is absent */
        System.out.println(search + " is not present in array.");
    }
}

Answer 1

...
System.out.println("Enter value to find");
search = in.nextInt();

boolean exists = false;

for (c = 0; c < n; c++)
{
  if (array[c] == search)     /* Searching element is present */
  {
     System.out.println(search + " is present at location " + (c + 1) + ".");
     exists = true;
  }
}

if (!exists)  /* Searching element is absent */
  System.out.println(search + " is not present in array.");

您需要删除break;语句。否则，一旦找到第一个值，循环就会中断，并且永远不会到达下一个匹配项。