Question

我想创建一个在排序数组中搜索线性的程序，并可以输出找到搜索项的不同位置。目前我的程序只输出找到搜索项的第一个位置，所以这里是我的程序现在所做的一个例子：

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

现在问题是3位于第2和第3位，这就是我想在程序中编辑但我不知道该怎么做。

这是我的程序代码：

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，循环就会被破坏，＆amp;永远不会达到下一场比赛。

Answer 2

此代码可能适合您。

    for (int i = first; i <= last; i++)
    {
        System.out.println(search + " is present at location " + (i + 1) + ".");
    }

现在索引“first”和“last”（包括它们）之间的元素包含搜索到的数字。因此，您基本上搜索与您要查找的第一个和最后一个元素将其写入控制台的示例：

/app/
    /controllers/
    /views/
    /lib/
    /public/
    .htaccess

Answer 3

import java.util.Scanner;

public class BinarySearch {
    /*
     * A binary search or half-interval search algorithm finds the position of a
     * specified value (the input "key") within a sorted array. Binary search
     * requires a sorted collection. Also, binary searching can only be applied
     * to a collection that allows random access (indexing).
     * 
     * Worst case performance: O(log n)
     * 
     * Best case performance: O(1)
     */

    public static int binSearch(int a[], int key) {
        int start = 0;
        int end = a.length - 1;

        while (start <= end) {
            int mid = (start + end) / 2;

            if (key == a[mid]) {
                return mid;
            }
            if (key < a[mid]) {
                end = mid - 1;
            } else {
                start = mid + 1;

            }

        }
        return -1;

    }

    public static void main(String arg[]) {

        BinarySearch bi = new BinarySearch();
        int[] arr = { 2, 4, 6, 8, 10, 12, 14, 16 };
        System.out.println("Please Key to be search");
        Scanner sc = new Scanner(System.in);
        int input = Integer.parseInt(sc.nextLine());

        if (bi.binSearch(arr, input) != -1) {
            System.out.println(input + ": " + " Search found at "
                    + bi.binSearch(arr, input) + " " + "Position");
        } else {
            System.out.println(input + ": " + " Search Result not found ");
        }

    }
}

Answer 4

使用 ArrayList 存储位置： -

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

类LinearSearchArray1 {

public static void main(String args[]){

    int c, n, search, array[];
    boolean searchStatus=false;

    ArrayList<Integer> al=new ArrayList<Integer>();

    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 */
        {
         al.add(c+1);
         searchStatus=true;  
    }
}
if (searchStatus==false)  /* Searching element is absent */
    System.out.println(search + " is not present in array.");
else {
    System.out.print(search+ " is present in array at location ");
    for(Integer i:al) 
        System.out.print(i+",");
}
}

}

排序数组中的线性搜索 - Java

4 个答案: