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    <p>If</p> <ul> <li>by "modification is not allowed" it was meant "you may change but in the end they should be restored", and</li> <li>we could iterate the lists exactly <em>twice</em></li> </ul> <p>the following algorithm would be the solution.</p> <p>First, the numbers. Assume the first list is of length <code>a+c</code> and the second one is of length <code>b+c</code>, where <code>c</code> is the length of their common "tail" (after the mergepoint). Let's denote them as follows:</p> <pre><code>x = a+c y = b+c </code></pre> <p>Since we don't know the length, we will calculate <code>x</code> and <code>y</code> without additional iterations; you'll see how.</p> <p>Then, we iterate each list and reverse them while iterating! If both iterators reach the merge point at the same time, then we find it out by mere comparing. Otherwise, one pointer will reach the merge point before the other one.</p> <p>After that, when the other iterator reaches the merge point, it won't proceed to the common tail. Instead will go back to the former beginning of the list that had reached merge-point before! So, before it reaches the end of the changed list (i.e. the former beginning of the other list), he will make <code>a+b+1</code> iterations total. Let's call it <code>z+1</code>.</p> <p>The pointer that reached the merge-point first, will keep iterating, until reaches the end of the list. The number of iterations it made should be calculated and is equal to <code>x</code>.</p> <p>Then, this pointer iterates back and reverses the lists again. But now it won't go back to the beginning of the list it originally started from! Instead, it will go to the beginning of the other list! The number of iterations it made should be calculated and equal to <code>y</code>.</p> <p>So we know the following numbers:</p> <pre><code>x = a+c y = b+c z = a+b </code></pre> <p>From which we determine that</p> <pre><code>a = (+x-y+z)/2 b = (-x+y+z)/2 c = (+x+y-z)/2 </code></pre> <p>Which solves the problem.</p>
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    1. POCheck if two linked lists merge. If so, where?
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    1. COComment to question states modification of list not allowed!
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    2. COI like this answer (very creative). The only problem I have with it is that it assumes you know the length of both lists.
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    3. COyou cannot modify list, and we don't know the length--these are the constraints...any how, thanks for a creative answer.
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