Approach #1 Recursion [Accepted]

Intuition

We can recursively define the result of a merge operation on two lists as\nthe following (avoiding the corner case logic surrounding empty lists):

\n $$\n\\left\\{\n\\begin{array}{ll}\n list1[0] + merge(list1[1:], list2) & list1[0] < list2[0] \\\\\n list2[0] + merge(list1, list2[1:]) & otherwise\n\\end{array} \n\\right.\n$$ \n

Namely, the smaller of the two lists\' heads plus the result of a merge on\nthe rest of the elements.

Algorithm

We model the above recurrence directly, first accounting for edge cases.\nSpecifically, if either of l1 or l2 is initially null, there is no\nmerge to perform, so we simply return the non-null list. Otherwise, we\ndetermine which of l1 and l2 has a smaller head, and recursively set the\nnext value for that head to the next merge result. Given that both lists\nare null-terminated, the recursion will eventually terminate.

Complexity Analysis

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  Time complexity : $$O(n + m)$$ \n

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  Because each recursive call increments the pointer to l1 or l2 by one\n(approaching the dangling null at the end of each list), there will be\nexactly one call to mergeTwoLists per element in each list. Therefore,\nthe time complexity is linear in the combined size of the lists.

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  Space complexity : $$O(n + m)$$ \n

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  The first call to mergeTwoLists does not return until the ends of both\nl1 and l2 have been reached, so $$n + m$$ stack frames consume\n $$O(n+m)$$ space.

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Approach #2 Iteration [Accepted]

Intuition

We can achieve the same idea via iteration by assuming that l1 is entirely\nless than l2 and processing the elements one-by-one, inserting elements of\nl2 in the necessary places in l1.

Algorithm

First, we set up a false "prehead" node that allows us to easily return the\nhead of the merged list later. We also maintain a prev pointer, which\npoints to the current node for which we are considering adjusting its next\npointer. Then, we do the following until at least one of l1 and l2 points\nto null: if the value at l1 is less than or equal to the value at l2,\nthen we connect l1 to the previous node and increment l1. Otherwise, we\ndo the same, but for l2. Then, regardless of which list we connected, we\nincrement prev to keep it one step behind one of our list heads.

After the loop terminates, at most one of l1 and l2 is non-null.\nTherefore (because the input lists were in sorted order), if either list is\nnon-null, it contains only elements greater than all of the\npreviously-merged elements. This means that we can simply connect the\nnon-null list to the merged list and return it.

To see this in action on an example, check out the animation below:

!?!../Documents/21_Merge_Two_Sorted_Lists.json:1280,720!?!

Complexity Analysis

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  Time complexity : $$O(n + m)$$ \n

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  Because exactly one of l1 and l2 is incremented on each loop\niteration, the while loop runs for a number of iterations equal to the\nsum of the lengths of the two lists. All other work is constant, so the\noverall complexity is linear.

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  Space complexity : $$O(1)$$ \n

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  The iterative approach only allocates a few pointers, so it has a\nconstant overall memory footprint.

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Analysis and recursive solution written by: @emptyset

Iterative solution written by: @1337c0d3r