21. Merge Two Sorted Lists || 2 Approach || LeetCode Placement Series || Explained in HINDI

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Approach 1:-
Base Cases:
If list1 is null, return list2 because merging with an empty list returns the other list.
If list2 is null, return list1 for the same reason.
Recursive Merge:
Compare the values of the current nodes of list1 and list2.
If the value of list1 is less than or equal to the value of list2, set list1.next to the result of merging the next node of list1 with list2. Return list1 as the head of the merged list.
Otherwise, set list2.next to the result of merging list1 with the next node of list2. Return list2 as the head of the merged list.
This approach recursively breaks down the problem, merging one node at a time based on the node values, until all nodes are merged.
Time Complexity
O(n + m): The function processes each node exactly once, where n and m are the lengths of the two linked lists.
Space Complexity
O(n + m): The recursion depth will be at most n + m, where n and m are the lengths of the two linked lists. Each recursive call adds a new frame to the stack, leading to linear space complexity in the worst case.
Approach 2:-
Initialization:
Check if either list1 or list2 is null. If so, return the non-null list because merging with an empty list results in the other list.
Setting the Starting Point:
Compare the first nodes of list1 and list2.
Initialize ptr with the smaller node and update the corresponding list to its next node.
Iterative Merging:
Use a while loop to continue merging nodes until one of the lists is exhausted.
At each step, compare the current nodes of list1 and list2.
Append the smaller node to the merged list and move to the next node in that list.
Update the curr pointer to the latest node added to the merged list.
Appending Remaining Nodes:
After the loop, if one list is exhausted, append the remaining nodes of the other list to the merged list.
Time Complexity
O(n + m): The function processes each node exactly once, where n and m are the lengths of the two linked lists.
Space Complexity
O(1): The iterative approach does not use additional space beyond the pointers used for traversal.
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21. Merge Two Sorted Lists
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Code link :- leetcode.com/problems/merge-two-sorted-lists/solutions/5551586/java-solution-explained-in-hindi/