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copied!<p>You can do better without list reversal. WLOG I'll assume that both lists have equal length (prepend with 0 if necessary).</p> <p>Start the addition from left to right (from most significant to least significant digit). You have three cases, depending of the sum of two digits:</p> <ol> <li>= 9: keep the nine and increase a <code>counter</code></li> <li>< 9: write <code>counter</code> x nine, write sum, reset <code>counter</code></li> <li><blockquote> <p>9: increase last digit, write <code>counter</code> x zero, write sum (modulo 10), reset counter</p> </blockquote></li> </ol> <p>I'll work on the following example:</p> <pre><code>2 568 794 + 1 438 204 --------- = 4 006 998 </code></pre> <ol> <li>Add <code>2 + 1 = 3</code>: case 3. <ul> <li><code>list = (3)</code>, <code>counter = 0</code></li> </ul></li> <li>Add <code>5 + 4 = 9</code>: case 1 <ul> <li><code>list = (3)</code>, <code>counter = 1</code></li> </ul></li> <li>Add <code>6 + 4 = 9</code>: case 1 <ul> <li><code>list = (3)</code>, <code>counter = 2</code></li> </ul></li> <li>Add <code>8 + 8 = 16</code>: case 3 <ul> <li><code>list = (4, 0, 0, 6)</code>, <code>counter = 0</code></li> </ul></li> <li>Add <code>7 + 2 = 9</code>: case 1 <ul> <li><code>list = (4, 0, 0, 6)</code>, <code>counter = 1</code></li> </ul></li> <li>Add <code>9 + 0 = 9</code>: case 1 <ul> <li><code>list = (4, 0, 0, 6)</code>, <code>counter = 2</code></li> </ul></li> <li>Add <code>4 + 4 = 8</code>: case 2 <ul> <li><code>list = (4, 0, 0, 6, 9, 9, 8)</code>, <code>counter = 0</code></li> </ul></li> </ol>

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