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Problem Solution class Node: def __init__(self, val): self.val = val self.next = None class MyLinkedList: def __init__(self): self.head = None self.size = 0 def get(self, index: int) -> int: if self.head==None or index None: curr = self.head if not curr: self.head = Node..

Problem Solution Ignore the random pointers for now. We need a way to make the next property point to a new node. First, create new nodes with original values in the linked list and plug them after the node of the original value. Then we can work through the interweaved list and take care of random pointers for the new nodes. Lastly, break the next links so that old nodes point to old nodes and ..

Problem Solution My initial approach was to compare the head nodes from each list, put the smallest onto the result linked list, and proceed the pointer that was on the smallest node. The time complexity for this approach would be O(kN) where k is the number of elements in lists and N is the number of nodes in the final linked list. I did not find a clean way to implement this. Solution using me..

Problem Solution We can easily see that this problem is just elementary school math, except that the digits are divided into nodes in the linked lists. Turn each list into an integer, then add the two numbers up. Finally, turn the sum into a linked list by iterating through the digits. Time Complexity O(N1+N2), where N1 and N2 are the numbers of elements in each input list. Space Complexity O(1)..

Problem Solution Use two pointers that move with different speeds. One moves forward by steps of 1, and the other moves by steps of 2. If the faster pointer reaches the end of the list, there is no loop in the list. If there is a loop in the list, the two pointers will intersect eventually. Return true when that happens. Check for edge cases where the list is empty or there are less than 3 eleme..

Problem Solution The base idea is to go through both lists at the same time, add each nodes, and append it to a result linked list. Make a variable to keep track of the carry for each sum. When making a result linked list, create two pointers for it. One pointer will always point to the head of the list. Another will go forward through the list as the code add elements to the linked list. In eac..