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16508133
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2013-05-12T13:59:24.120
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2013-05-12T14:21:20.747
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2013-05-12T14:21:20.747
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Assuming you're looking for which columns contain ones and now how many ones each column contains, looping through them seems like the best idea. If you implement your loops with some short-circuit logic, then you'll get a better average running time. Something like: <pre><code>for (int column = 0; column < width; column++) { for (int row = 0; row < height; row++) { if (array[column][row] == 1) { list.append(column); break; // move on to the next column because we don't care what's // left in this column as we already found our '1' </code></pre> With this code, you'll get a worst-case (if every <code>bit</code> is <code>0</code>) running time of <code>n</code> (<code>n</code> being the total number of <code>bits</code>) which is pretty good. But unless you're extremely unlucky, you'll get a much better running time due to the short-circuit logic. The above works for arrays of bits, however, <code>BitSets</code> have some useful functionality of their own. A <code>BitSet</code> contains the functions: <code>length()</code> which returns the highest index with a set bit + 1 (or <code>0</code> if there are no set bits) and <code>nextSetBit(index)</code> which returns the index of the next set bit from <code>index -> end</code> inclusive. So you're code could easily be something like: <pre><code>int index = 0; while (index < BitSet.length()) { index = BitSet.nextSetBit(index); list.append(index); index++; } </code></pre>
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1. This table or related slice is empty.
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1. POEfficient way of creating a bit matrix and scan set columns
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1. USTheMerovingian
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1. USTheMerovingian
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1. POEfficient way of creating a bit matrix and scan set columns
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