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copied!<p>What did you try? </p> <p>The idea, given the problem you stated (which specifies we must use recursion) is simple: for each item that you can take, see if it's better to take it or not. So there are only two possible path:</p> <ol> <li>you take the item</li> <li>you don't take it</li> </ol> <p>When you take the item, you remove it from your list and you decrease the capacity by the weight of the item.</p> <p>When you don't take the item, you remove if from you list but you do not decrease the capacity.</p> <p>Sometimes it helps to print what the recursive calls may look like. In this case, it could look like this:</p> <pre><code>Calling 11 8 7 6 5 with cap: 20 +Calling 8 7 6 5 with cap: 20 | Calling 7 6 5 with cap: 20 | Calling 6 5 with cap: 20 | Calling 5 with cap: 20 | Result: 5 | Calling 5 with cap: 14 | Result: 5 | Result: 11 | Calling 6 5 with cap: 13 | Calling 5 with cap: 13 | Result: 5 | Calling 5 with cap: 7 | Result: 5 | Result: 11 | Result: 18 | Calling 7 6 5 with cap: 12 | Calling 6 5 with cap: 12 | Calling 5 with cap: 12 | Result: 5 | Calling 5 with cap: 6 | Result: 5 | Result: 11 | Calling 6 5 with cap: 5 | Calling 5 with cap: 5 | Result: 5 | Result: 5 | Result: 12 +Result: 20 Calling 8 7 6 5 with cap: 9 Calling 7 6 5 with cap: 9 Calling 6 5 with cap: 9 Calling 5 with cap: 9 Result: 5 Calling 5 with cap: 3 Result: 0 Result: 6 Calling 6 5 with cap: 2 Calling 5 with cap: 2 Result: 0 Result: 0 Result: 7 Calling 7 6 5 with cap: 1 Calling 6 5 with cap: 1 Calling 5 with cap: 1 Result: 0 Result: 0 Result: 0 Result: 8 Result: 20 </code></pre> <p>I did on purpose show the call to [8 7 6 5] with a capacity of 20, which gives a result of 20 (8 + 7 + 5).</p> <p>Note that [8 7 6 5] is called twice: once with a capacity of 20 (because we didn't take 11) and once with a capacity of 9 (because with did take 11).</p> <p>So the path to the solution:</p> <p>11 not taken, calling [8 7 6 5] with a capacity of 20</p> <p>8 taken, calling [7 6 5] with a capacity of 12 (20 - 8)</p> <p>7 taken, calling [6 5] with a capacity of 5 (12 - 7)</p> <p>6 not taken, calling [5] with a capacity of 5</p> <p>5 taken, we're at zero.</p> <p>The actual method in Java can fit in very few lines of code.</p> <p>Since this is obviously homework, I'll just help you with a skeleton:</p> <pre><code>private int ukp( final int[] ar, final int cap ) { if ( ar.length == 1 ) { return ar[0] <= cap ? ar[0] : 0; } else { final int[] nar = new int[ar.length-1]; System.arraycopy(ar, 1, nar, 0, nar.length); fint int item = ar[0]; if ( item < cap ) { final int left = ... // fill me: we're not taking the item final int took = ... // fill me: we're taking the item return Math.max(took,left); } else { return ... // fill me: we're not taking the item } } } </code></pre> <p>I did copy the array to a new array, which is less efficient (but anyway recursion is not the way to go here if you seek performance), but more "functional".</p>

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