LeetCode #1290 — EASY

Convert Binary Number in a Linked List to Integer

Build confidence with an intuition-first walkthrough focused on linked list fundamentals.

The Problem

Problem Statement

Given head which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1. The linked list holds the binary representation of a number.

Return the decimal value of the number in the linked list.

The most significant bit is at the head of the linked list.

Example 1:

Input: head = [1,0,1]
Output: 5
Explanation: (101) in base 2 = (5) in base 10

Example 2:

Input: head = [0]
Output: 0

Constraints:

The Linked List is not empty.
Number of nodes will not exceed 30.
Each node's value is either 0 or 1.

Step 01

Brute Force Baseline

Problem summary: Given head which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1. The linked list holds the binary representation of a number. Return the decimal value of the number in the linked list. The most significant bit is at the head of the linked list.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Linked List · Math

Example 1

[1,0,1]

Example 2

[0]

Step 02

Core Insight

What unlocks the optimal approach

Traverse the linked list and store all values in a string or array. convert the values obtained to decimal value.
You can solve the problem in O(1) memory using bits operation. use shift left operation ( << ) and or operation ( | ) to get the decimal value in one operation.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1290: Convert Binary Number in a Linked List to Integer
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public int getDecimalValue(ListNode head) {
        int ans = 0;
        for (; head != null; head = head.next) {
            ans = ans << 1 | head.val;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1290: Convert Binary Number in a Linked List to Integer
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func getDecimalValue(head *ListNode) (ans int) {
	for ; head != nil; head = head.Next {
		ans = ans<<1 | head.Val
	}
	return
}

# Accepted solution for LeetCode #1290: Convert Binary Number in a Linked List to Integer
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def getDecimalValue(self, head: ListNode) -> int:
        ans = 0
        while head:
            ans = ans << 1 | head.val
            head = head.next
        return ans

// Accepted solution for LeetCode #1290: Convert Binary Number in a Linked List to Integer
// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn get_decimal_value(mut head: Option<Box<ListNode>>) -> i32 {
        let mut ans = 0;
        while let Some(node) = head {
            ans = (ans << 1) | node.val;
            head = node.next;
        }
        ans
    }
}

// Accepted solution for LeetCode #1290: Convert Binary Number in a Linked List to Integer
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function getDecimalValue(head: ListNode | null): number {
    let ans = 0;
    for (; head; head = head.next) {
        ans = (ans << 1) | head.val;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

COPY TO ARRAY

O(n) time

O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS

O(n) time

O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.