Teaching Kids Programming: Videos on Data Structures and Algorithms
ou are given a singly linked list node containing positive integers. Return the same linked list where every node’s next points to the node val nodes ahead. If there’s no such node, next should point to null.
Constraints
n ≤ 100,000 where n is the number of nodes in node
Example 1
Input
node = [2, 1, 4, 1]Output
[2, 4]Explanation
Note that 2’s next is 2 node ahead. 4’s next is out of bounds, so it’s set to null.
Linked List Jumps via Recursion
We can solve this one step at a time. We can find out the destination node from the current node with n jumps – and break when we reach the end. Then, we assign the next pointer of the current node to the destination node – which we can recursively jump.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | # class LLNode: # def __init__(self, val, next=None): # self.val = val # self.next = next class Solution: def jump(self, node): if not node: return node head = node for _ in range(head.val): head = head.next if not head: break node.next = self.jump(head) return node |
# class LLNode:
# def __init__(self, val, next=None):
# self.val = val
# self.next = next
class Solution:
def jump(self, node):
if not node:
return node
head = node
for _ in range(head.val):
head = head.next
if not head:
break
node.next = self.jump(head)
return nodeSee also: Recursive Algorithm to Compute the Linked List Jumps
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