Code a multiplication table.

Question

Anónimo · Accepted Answer

public static void main(String []args){
        for(int i = 1; i<=9; i++){
            for(int j=1; j<=9; j++ ){
                System.out.printf("%4d", i*j);
            }
            System.out.println();
        }
     }

Yates · Answer

// An alternate O(n^2) solution. I feel like you could make this O(n log n) somehow if you stored previous multiplication pairs in some data structure.

int[] numbers = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
        int colCount = 1;
        int rowCount = 1;
        int innerCount = 0;

while (rowCount != numbers.length + 1) {
            System.out.printf("%4d", numbers[innerCount++] * colCount);

if (innerCount == numbers.length) {
                innerCount = 0;
                colCount++;
                rowCount++;
                System.out.println("");
            }
        }

Yates · Answer

Adding to above answer. You can think of it like storing results of multiplying:
1 -> 9
2 -> 9
3 -> 9
...
8-> 9
9 -> 9 
So that you're only computing the values once. So if you needed 9 * 2 (larger times smaller), you would look up the where computations of 2 (the smaller number) are stored and grab the value of 9. A 2D jagged array might could work for this.

Fidelity Investments

Pregunta de entrevista de Fidelity Investments

Respuestas de entrevistas

Empresas seguidas

Búsquedas de empleo