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e73ac6115f | ||
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959c579550 |
@@ -1,61 +1,65 @@
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public class MainExercises
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{
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/*
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you should create a triangle with "*" and return a two-dimensional array of characters based on that
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the triangle's area is empty, which means some characters should be " "
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example 1, input = 3:
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*
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**
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***
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example 2, input = 5:
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*
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**
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* *
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* *
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*****
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the output has to be a two-dimensional array of characters, so don't just print the triangle!
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*/
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public char[][] generateTriangle(int n) {
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char[][] result = new char[n][n];
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for (int i = 0; i < n; i++)
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{
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if (i == n-1)
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{
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for (int j = 0; j <= i; j++)
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{
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result[i][j] = '*';
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}
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}
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else
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{
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result[i][0] = '*';
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result[i][i] = '*';
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}
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}
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// todo
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return null;
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return result;
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}
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/*
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SPIRAL TRAVERSAL OF A RECTANGULAR MATRIX
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Given a rectangular matrix (2D array) of integers, this method traverses
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it in a spiral order (clockwise from outside to inside) and returns
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the elements as a 1D array.
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EXAMPLE:
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Input matrix:
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1 2 3
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4 5 6
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7 8 9
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Spiral order: start at top-left (1), go right →, then down ↓,
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then left ←, then up ↑, then repeat inward.
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Result: {1, 2, 3, 6, 9, 8, 7, 4, 5}
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so you should walk in that matrix in a curl and then add the numbers in order you've seen them in a 1D array
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RECTANGULAR MATRIX ASSUMPTION:
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This method assumes the input matrix is RECTANGULAR (all rows have
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the same number of columns). In Java, we can verify this because
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2D arrays might be jagged (rows of different lengths).
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IMPORTANT: In Java, we do NOT need to pass rows and cols!
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The 2D array 'matrix' knows its own dimensions:
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- Number of rows: matrix.length
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- Number of columns: matrix[0].length (if rectangular)
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*/
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public int[] spiralTraversal(int[][] matrix) {
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int rows = matrix.length, cols = matrix[0].length;
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int elementCount = rows * cols, curElement = 0;
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int[] result = new int[elementCount];
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for (int leyer = 0; curElement < elementCount; leyer++) {
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int curi = leyer, maxi = rows - 1 - leyer;
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int curj = leyer, maxj = cols - 1 - leyer;
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for (;curj < maxj && curElement < elementCount; curj++, curElement++)
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{
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result[curElement] = matrix[curi][curj];
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}
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for (;curi < maxi && curElement < elementCount; curi++, curElement++)
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{
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result[curElement] = matrix[curi][curj];
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}
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for (;curj > leyer && curElement < elementCount; curj--, curElement++)
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{
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result[curElement] = matrix[curi][curj];
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}
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for (;curi > leyer && curElement < elementCount; curi--, curElement++)
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{
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result[curElement] = matrix[curi][curj];
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}
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if (curi == maxi || curj == maxj) {
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if (curi == maxi && curj == maxj) result[curElement] = matrix[curi][curj];
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break;
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}
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}
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return result;
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// todo
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return null;
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}
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/*
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