How do I fill an array with consecutive numbers

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感情败类
感情败类 2020-12-20 12:29

I would like to fill an array using consecutive integers. I have created an array that contains as much indexes as the user enters:

Scanner in = new Scanner(         


        
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  • 2020-12-20 13:05

    One more thing. If I want to do the same with reverse:

    int[] array = new int[5];
            for(int i = 5; i>0;i--) {
                array[i-1]= i;
            }
            System.out.println(Arrays.toString(array));
    }
    

    I got the normal order again..

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  • 2020-12-20 13:13

    Since Java 8

    //                               v end, exclusive
    int[] array = IntStream.range(1, numOfValues + 1).toArray();
    //                            ^ start, inclusive
    

    The range is in increments of 1. The javadoc is here.

    Or use rangeClosed

    //                                     v end, inclusive
    int[] array = IntStream.rangeClosed(1, numOfValues).toArray();
    //                                  ^ start, inclusive
    
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  • 2020-12-20 13:13
    for(int i=0; i<array.length; i++)
    {
        array[i] = i+1;
    }
    
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  • 2020-12-20 13:15

    The simple way is:

    int[] array = new int[NumOfValues];
    for(int k = 0; k < array.length; k++)
        array[k] = k + 1;
    
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  • 2020-12-20 13:20
    Scanner in = new Scanner(System.in);
    int numOfValues = in.nextInt();
    
    int[] array = new int[numOfValues];
    
    int add = 0;
    
    for (int i = 0; i < array.length; i++) {
    
        array[i] = 1 + add;
    
        add++;
    
        System.out.println(array[i]);
    
    }
    
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  • 2020-12-20 13:23

    You now have an empty array

    So you need to iterate over each position (0 to size-1) placing the next number into the array.

    for(int x=0; x<NumOfValues; x++){ // this will iterate over each position
         array[x] = x+1; // this will put each integer value into the array starting with 1
    }
    
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