在数组中搜索一个递增的整数序列

Question

Here's the task:

Create a program that looks through a 200 -element array of randomly-generated integers with values between 25 and 225 . The program should look for n consecutive indices that hold non-decreasing values - for example:

28, 57, 88, 153, 201 is considered a success

28, 57, 32, 153, 201 is considered a failure.

Output should look something like:

An increasing consecutive run of 5 elements found at:

Index 41 contains 28
Index 42 contains 57
Index 43 contains 88
Index 44 contains 153
Index 45 contains 201

There are many ways to go about this task. I've included a working step-by-step solution below.

Answer 1

SOLUTION

First, apply the given information (lower bound, upper bound, array size).

    int min = 25;
    int max = 225;
    int set[] = new int[200];

Set n to an arbitrary value. I will use 5 for demonstration purposes.

    int n = 5;

Create a counter variable to record the number of consecutive integers found. Further on we will add logic to reset the counter every time a sequence of numbers does not work.

    int counter = 0;

Start iterating. Create a for loop to fill the array with '200' elements between the specified bounds.

    for(int i = 0; i < set.length; i++){
        set[i] = (int)(Math.random() * (max - min + 1) + min);

Print out the contents of every index.

        System.out.println((i + 1) +") " + "Index " + i + " contains " + set[i]);

The first if statement: this will increment 'counter' by 1 every time a sequence fits the pattern, but resets it back to 0 when the comparison fails.

   if(i > 0){   
        if(set[i] >= set[i-1]){counter++;}
        else{counter = 0;}  
    }// End if

Now create a while statement that activates when the value of counter becomes equal to the value of n . This will print out the statement An increasing consecutive run of 5 elements found at: and then lead to a for loop.

    while(counter == n){

           System.out.println("An increasing consecutive run of " + n + " elements found at:");

Now set up this loop to simply print out the n lines that were successful.

    for(int j = i - n; j < i; j++){
        System.out.println("Index " + j + " contains " + set[j]);
    } //End for

Tell the code to stop running once all conditions are met.

       return;

Just close the while and for loops and the program is now complete.

        } //End while

    } //End for