Finding the minimum value of int numbers in an arr

问题:

I'm trying to find the minimum value of numbers in an array but it doesn't always work out properly. This is the code I wrote:

        for (int i=0; i < arr.length; i++ ) {
        min = arr[i];
        for (j=0; j < arr.length; j++) {
        if (arr[j] < arr[0]) {
            min = arr[j];
         }
        }
    }   

Can someone correct me please?

回答1:

There's no need for the outer loop, it only runs once and you don't use i anyway. why do you have it?

For the inner loop, you need to compare against the minimum value. Right now you are comparing it against the first element in the array, which is not necessarily the minimum value.

min = arr[0];
for (j=0; j < arr.length; j++) {
    if (arr[j] < min) {  //<---fix is here
        min = arr[j];
    }
}

Also you could start the loop at 1, since you don't need to compare arr[0] against itself (it was just assigned to min)

回答2:

int min = arr[0];
for(int num : arr) {
    if (num < min){
        min = num;
    }
}

min now contains minimum value.

回答3:

A way to do it would be using the java.util.Arrays class:

Example:

public class ArraySort {
public static void main(String[] args) {
    int[] array = {12, 4, 6, 1, 56, 21, 77};
    Arrays.sort(array);
    System.out.println(array[0]);
}
}

From the Java doc, Arrays.sort(int[]) sort the specified array into ascending numerical order.

So the output here prints 1.

回答4:

One option is sort your array and get the first element:

import java.util.Arrays;

...

int ints[] = {30,5,7,4,10};
Arrays.sort(ints);

int min = ints[0];
int max = ints[ints.length - 1];

回答5:

If arr is an array of non-primitive numerics, I'd recommend

java.util.Collections.min(java.util.Arrays.asList(arr));

as that will be simpler to maintain. There's some justification in reverting to hand-coding if you need to pull the minimum and maximum out at the same time but I'd advise against hand-coding loops if there's a library function available.

In any case, you ought to check arr != null and the existence of a zeroth element.

回答6:

You are checking the first element in each iteration, you basically need to check minimum value

if (arr[j] < min) {
  min = arr[j];
}

回答7:

Here is a general algorithm for doing this. You can write the code for it.

Store the first item in the array as the current minimum.

Loop through the array, starting at the second item (index 1).

For each iteration of the array, check if the current item is less than the minimum. If it is, store it as the new minimum.

Once the loop ends, you have the minimum!

回答8:

TRY this:

int min = arr[0];
    for(int j=1;j<arr.length;j++)
    {
            if(min>arr[j])
            {
                min= arr[j];
            }
    }
    System.out.println("min no is "+min);

回答9:

      int min=0; 
        for (int i = 0; i < array.length; i++) {
        if (min > array[i]) {
        min = array[i];
        }
}

    System.out.println(min);

simple way to get MAX and MIN

To get MIN

System.out.println(getMinValue(your array)); 

and for MAX

System.out.println(getMaxValue(your array));