Another simple question, doing some bit arithmetic. We traverse the linked list from the head to the tail, and for each node, we shift the current value to the left by 1 and or the value of the current node.
The solution is as follows:
class Solution:
def getDecimalValue(self, head: ListNode) -> int:
num = head.val
while head.next:
num = (num << 1) | head.next.val
head = head.next
return num_ Time Complexity:
O(n) - We traverse all nodes within the linked list.
_ Space Complexity:
O(1) - We maintain pointers that occupy constant space.