how to get the minimum,maximum value of an array? [duplicate]

Answer 1

int minIndex = list.indexOf(Collections.min(list));

or

public class MinMaxValue {

    public static void main(String[] args) {
        char[] a = {'3', '5', '1', '4', '2'};

        List b = Arrays.asList(ArrayUtils.toObject(a));

        System.out.println(Collections.min(b));
        System.out.println(Collections.max(b));
   }
}

Answer 2

In your code remove the following part:

(min = max = convertedValues[0];)

and initialize min to 1:

(int min = 1) 

Answer 3

public static int[] getMinMax(int[] array) {

    int min = Integer.MAX_VALUE; //or -1
    int max = Integer.MIN_VALUE; //or -1
    for(int i : array) { //if you need to know the index, use int (i=0;i<array.length;i++) instead
        if(i < min) min = i; //or if(min == -1 || array[i] < array[min]) min = i; 
        if(i > max) max = i; //or if(max == -1 || array[i] > array[max]) max = i;
    }
    return new int[] {min, max};
}

Sorting takes at least O(n log(n)), n being the amount of elements in the array. If you simply look at every element in the array, finding the minimum and maximum element is in O(n). For big arrays, that's a lot faster.

Answer 4

int[] convertedValues = new int[10];
int max = convertedValues[0];

for (int i = 1; i < convertedValues.length; i++) {
    if (convertedValues[i] > max) {
        max = convertedValues[i];
    }
}

Similarly find for the minimum value by changing lesser symbol.

Answer 5

You could sort the array and get the positions 0 and length-1:

Arrays.sort(convertedValues);

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

Arrays#sort(int[]):

Sorts the specified array of ints into ascending numerical order.

So, after sorting, the first element is the minimum, the last is the maximum.

Answer 6

I think you are just missing one check .. You should define the min and max variables as -1 and add following check.

if (min == -1) {
    min = convertedValues[i];
}