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dd399fd00a
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dd399fd00a | ||
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95028697fd |
@@ -1,25 +1,35 @@
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import java.util.ArrayList;
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public class MainExercises
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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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char[][] result = new char[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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char[] line = new char[i+1];
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if (i == n-1 || i == 0)
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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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line[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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line[0] = '*';
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line[i] = '*';
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for (int j = i - 1; j > 0; j--) {
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line[j] = ' ';
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}
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}
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result[i] = line;
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}
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return result;
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}
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@@ -58,37 +68,36 @@ public class MainExercises
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return result;
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}
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/*
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integer partitioning is a combinatorics problem in discreet maths
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the problem is to generate sum numbers which their summation is the input number
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e.g. 1 -> all partitions of integer 3 are:
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3
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2, 1
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1, 1, 1
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e.g. 2 -> for number 4 goes as:
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4
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3, 1
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2, 2
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2, 1, 1
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1, 1, 1, 1
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Note: As you can see in the examples, we want to generate distinct partitions,
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which means 1,2 and 2,1 are not different — they count as the same combination.
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You should generate all partitions of the input number.
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Hint: You can determine the size and order of the arrays by finding the pattern
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of partitions and their count. Trust me, this one's fun and easy :)
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If you're familiar with Lists and ArrayLists, you can also edit the method's
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body to use them instead of arrays.
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*/
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public int[][] intPartitions(int n) {
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// todo
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return null;
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ArrayList<int[]> result = new ArrayList<>();
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ArrayList<Integer> curPartition = new ArrayList<>();
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fillPartition(n, n, curPartition, result);
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return result.toArray(new int[result.size()][]);
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}
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private void fillPartition(int max, int remaining, ArrayList<Integer> curPartition, ArrayList<int[]> result) {
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if (remaining == 0) {
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int size = curPartition.size();
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int[] savingArr = new int[size];
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for (int i = 0; i < size; i++) {
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savingArr[i] = curPartition.get(i);
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}
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result.add(savingArr);
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return;
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}
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for (int i = Math.min(max, remaining); i >= 1; i--) {
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curPartition.add(i);
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int newRemaining = remaining - i;
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fillPartition(i, newRemaining, curPartition, result);
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curPartition.removeLast();
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}
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}
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