使用Java方法查找数组中最小int的位置

Question

我编写了以下方法来查找数组中最小数字的位置/索引

private static int indexOfMin( int[] a, int cnt )
    {
        int loc = 0;//the variable that will hold the index position
        int min = a[loc];//the variable that will compare the value of loc against its location
        for (int i = 1; i < a.length; i++)
        {
            if (a[i] < min )//if value is less
            {
               min = a[i];// make value of i equal to min
               loc = i; loc takes on value of the index of the value of min
            }
       }
       return loc ;
    }

它不返回最小int的位置，而是返回最后一个int的位置。如何找到最小int的位置并将其返回到int loc

进一步编辑： 这是整个计划。另外两种方法我还在研究中。所以不要理他们

public static void main (String[] args) throws Exception
    {
        final int CAP = 20; // capacity
        Scanner infile = new Scanner( new File(args[0]) );
        int[] arr = new int[ CAP ];
        int count = 0;

        while ( count < arr.length && infile.hasNextInt() )
        {
            arr[ count ] = infile.nextInt();
            count++;
        }
        infile.close();
        printArray( arr, count );

        // this method is given to you as as. don't modity it
        int minVal = minOf( arr, count );
        System.out.println( "Smallest number is: " + minVal);

        // YOU MUST WRITE THE DEFINITION (code) BELOW MAIN FOR indexOfMin METHOD
        int indOfMin = indexOfMin( arr, count );
        System.out.println( "Smallest number is located at index position: " + indOfMin );

        // YOU MUST WRITE THE DEFINITION (code) BELOW MAIN FOR maxOf METHOD
        int maxVal = maxOf( arr, count );
        System.out.println( "Largest number is: " + maxVal);

        // YOU MUST WRITE THE DEFINITION (code) BELOW MAIN FOR indexOfMax METHOD
        int indOfMax = indexOfMax( arr, count );
        System.out.println( "Largest number is located at index position: " + indOfMax );

    } // END main

    // GIVEN AS IS - DO NOT MODIFY
    private static int minOf( int[] a, int cnt )
    {
        int min = a[0];
        for ( int i=0 ; i<cnt ; i++ )
        {
            if (a[i] < min)
                min = a[i];
        }
        return min;
    }

    // YOU WRITE  DEFINTION OF indexOfMin
    // returns the INDEX of the min value NOT the min value itself
    private static int indexOfMin( int[] a, int cnt )
    {
        int loc = 0;
        int min = a[loc];
        for (int i = 1; i < a.length; i++)
        {
            if (a[i] < min )
            {
               min = a[i];
               loc = i;
            }
       }
       return loc ;
    }

日期文件包含以下信息：

86 95 84 94 32 8 56 51 98 20 90 1 75 6 21

Answer 1

为什么不使用Arrays.sort()并返回数组的第一个元素

//Snippet
void minInArray(){
    int[] arr = new int[]{55,42,11,20,584,63,21,27,84,96,32,30};
    int[] cArr = Arrays.copyOf(arr, arr.length);

    Arrays.sort(cArr);
    int idx = Arrays.asList(arr).indexOf(cArr[0]);
    String s = new StringBuilder("The min value is: ")
               .append(cArr[0])
               .append(" located at index: ")
               .append(idx)
               .toString();
    System.out.println(s);
}

Answer 2

int[] arr = {1, 5, 3, 4, 0, 9};
System.out.println(indexOfMin(arr));

印刷品：

4

这是绝对正确的，因为数组的索引以0开头

Answer 3

第一种方法有效

你没有问题

您可能忘记了索引从0开始，因此将结果解释为错误

Answer 4

根据你的评论和编辑，我认为你想要

private static int indexOfMin(int[] a, int cnt) {
  int loc = 0;
  int min = a[loc];
  for (int i = 1; i < cnt; i++) {
    if (a[i] < min) {
      min = a[i];
      loc = i;
    }
  }
  return loc;
}

那么为了验证,

// this method is given to you as as. don't modity it
int minVal = minOf( arr, count );
System.out.println( "Smallest number is: " + minVal);

// YOU MUST WRITE THE DEFINITION (code) BELOW MAIN FOR indexOfMin METHOD
int indOfMin = indexOfMin( arr, count );
System.out.println( "Smallest number is located at index position: " + indOfMin);
if (arr[indOfMin] == minVal) {
  System.out.println("Min value passed");
} else {
  System.out.println("Min value failed");
}

Answer 5

如上所述，您的第一段代码工作正常

java中的数组从索引=0开始

从外观上看，cnt应该是您要传递的数据数组的长度，它可以用作.length的替代品（尽管传递额外数据是非常不必要的）

在第二段代码中，在完成整个循环（可能）之前返回一个索引。一旦到达第一个小于[0]的项并返回该位置（在本例中可能发生在索引3处），此操作将停止

为便于将来参考，了解您正在测试的数据将对我们有所帮助

Answer 6

第一个代码是正确的。您仍然可以删除从未使用过的cnt。第二个是绝对错误的

导入java。util。扫描仪

public class small {

private static int indexOfMin( int[] a ,int cnt)

{
   int loc=0;
    int min = a[0];
    for (int i = 1; i <cnt; i++)
    {
        if (a[i] < min )
        {
           min = a[i];
           loc = i;
        }
   }
   return loc ;
}
public static void main(String args[]){
    Scanner s= new Scanner(System.in);
    System.out.println("how many numbers do you want to enter:");
    int n=s.nextInt();
    int a[]=new int[n];
    System.out.println("enter the numbers:");
    for(int i=0;i<n;i++){
      a[i]=s.nextInt();
    }
    int p=indexOfMin(a,n);
    System.out.println((p+1)); //p+1 as we have taken location same as i which starts with 0
 }
}

您要输入多少个数字：7

输入数字： 34 45 33 234 123 116 555

最小no的位置为：3

我希望这对你有帮助