Posted on Nov 1, 2020 • Edited on Nov 16, 2020

Data Structures & Algorithms using Rust: Convert Binary Number in a Linked List to Integer

As part of Data Structures & Algorithms implementation using Rust, let's see how to solve Convert Binary Number in a Linked List to Integer problem.

Leetcode problem link: https://leetcode.com/problems/convert-binary-number-in-a-linked-list-to-integer/

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.

Instructions

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 Example 3: Input: head = [1] Output: 1 Example 4: Input: head = [1,0,0,1,0,0,1,1,1,0,0,0,0,0,0] Output: 18880 Example 5: Input: head = [0,0] 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.

Solution

Please find below the solution implemented using Rust

// 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(head: Option<Box<ListNode>>) -> i32 { let mut iter = head.iter(); let mut v = Vec::new(); loop { let n = iter.next(); match n { Some(x) => { v.push(x.val); iter = x.next.iter(); }, None => break }; } v.iter().fold(0, |acc, &b| acc*2 + b as i32) } }