LeetCode 1290: Convert Binary Number in a Linked List to Integer ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2025-07-14

Submissions: Accepted (first try)

Approach
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The linked list holds a binary number from the most significant bit to the least.

My Solution: store in an array, then compute backwards
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values = []
while head:
    values.append(head.val)
    head = head.next
res = 0
for i, bit in enumerate(values[::-1]):
    res += bit * (2 ** i)

Correct, but it uses an extra array and a reversal.

Better: compute while walking
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Starting from the most significant bit, shift the result left by one (multiply by 2) for each new bit, then add the bit. It is the same idea as decimal 123 = ((1 × 10) + 2) × 10 + 3:

class Solution:
    def getDecimalValue(self, head: Optional[ListNode]) -> int:
        res = 0
        while head:
            res = (res << 1) | head.val
            head = head.next
        return res

Complexity
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Time O(n), space O(1).

Mistakes to Avoid
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Accumulating 2 ** i from the head: the head is the most significant bit, not the least, so you would need the length or a reversal first.

Interview Explanation
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The list starts at the most significant bit, so I compute while walking: for each bit, shift the result left by one and add the bit; at the tail, that is the answer. No array or length is needed, so time is O(n) and space O(1).