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Thread: Second largest

  1. #1
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    Default Second largest

    Hi.

    I'm trying to find the largest and the second largest number in vector. But I can't.

    Can you help?

    public class Minmax{
      public static void main(String[] args) {
     
        //int min = Integer.MIN_VALUE;
        int max = Integer.MAX_VALUE;
        int max2 =Integer.MAX_VALUE;
     
     
        while (!StdIn.isEmpty()) {
          int N = StdIn.readInt();
          if (N > max){
            max2=max;//storing the old max
    	max = N;//the new max
    		}
     
          else if (N>max2 && N<max) {//if the number is between new max and the old max
            max2=N;
          }
    	}
    	StdOut.println(max2 + " " + max);
      }
    }
    Error message:
    C:\>java Minmax 2 22 5 55 6 66
    ^Z
    2147483647 2147483647

    C:\Users>java Minmax
    55 2 66 2 88 5 99
    ^Z
    2147483647 2147483647


  2. #2
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    Default Re: Second largest

    Remind me, what is StdIn? I think I remember it from a book I read a long time ago, but I've forgotten.

    It looks like you meant to use the command line input that is saved in the args[] array rather than StdIn. Is that right?

  3. #3
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    Default Re: Second largest

    StdIn is used in my textbook, Introduction to programming in java. It's a library thing for all kinds of inputs.

    Your right. But I think I can use it like that and press ctrl+enter

    --- Update ---

    This is the ''older'' version. As you can see it prints out the highest value twice.
    I'm trying to fix that.
    public class Minmax {
      public static void main(String[] args) {
     
        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        int max2 =Integer.MIN_VALUE;
     
        while (!StdIn.isEmpty()) {
          int N = StdIn.readInt();
          if (N > max) {
            max = N;
    	max2=max;
          }
          if (N < max && N>max2) {
            max2 = N;
          }
        }
        StdOut.println(max2 + " " + max);
      }
    }

    C:\java Minmax
    5
    55
    66
    55
    556
    5
    557
    ^Z
    557 557

  4. #4
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    Default Re: Second largest

    It's interesting that you changed the beginning values to MAX from MIN. If you want to find a large number from a collection, you start by comparing it to a smaller number, not a number that it couldn't possibly be larger than.

    I also don't see the purpose of your else statement. Try changing the initial values back to MIN, eliminate the else clause, and I think you'll be there.

  5. #5
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    Default Re: Second largest

    It did not work out for me.

    How about this. Find the three largest numbers and print them out.

    public class Minmax2 {
      public static void main(String[] args) {
     
        int max = Integer.MIN_VALUE;
        int max1= Integer.MIN_VALUE;
        int max2 =Integer.MIN_VALUE;
     
        while (!StdIn.isEmpty()) {
          int N = StdIn.readInt();
          if (max<N) {
            max2 = max1;
    		max1=max;
    		max=N;
    		}
    		if (max1<N) {
    			max2= max1;
    			max1 =N;
    			}
    			else if(max2<N){
    				max2=N;
    				}
        }
      StdOut.println(max2 + " " + max1 + " " + max);
    }}

    C:\>java Minmax2
    55
    66
    88
    -1
    66
    67
    68
    ^Z
    68 88 88

  6. #6
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    Default Re: Second largest

    You're right, I over simplified, but why did you change the problem?

    Here's my for loop using an array that contains the same numbers you used in the same order for the 2 largest numbers, not 3:
    int[] numbers = { 55, 66, 88, -1, 66, 67, 68 };
     
    for ( int i = 0 ; i < numbers.length ; i++ )
    {
        int N = numbers[i];
        if (N > max)
        {
            max2=max;//storing the old max
            max = N;//the new max
        }
     
        // find values between current max and max2 for new max2
        if ( N < max && N > max2 )
        {
            max2 = N;
        }
    }

  7. The Following User Says Thank You to GregBrannon For This Useful Post:

    JohnJohnson123 (October 18th, 2013)

  8. #7
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    Default Re: Second largest

    I thought it was more fun like this. I saw it on stackverflow.

    Thanks for the ongoing support.

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