/** 1290. Convert Binary Number in a Linked List to Integer https://leetcode.com/problems/convert-binary-number-in-a-linked-list-to-integer/ 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. 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: 1. The Linked List is not empty. 2. Number of nodes will not exceed 30. 3. Each node's value is either 0 or 1. */ // 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, } } } pub struct Solution {} impl Solution { /* solution: get all element of linked-list then calculate the sum of decimal, Time:O(n) */ pub fn get_decimal_value(head: Option<Box<ListNode>>) -> i32 { let mut result = 0; let mut size = 0; let mut current = head; let mut vector = Vec::new(); //keep doing {} while current can explain as Some(node) while let Some(node) = current { size += 1; vector.push(node.val); current = node.next } for item in vector { size -= 1; /* to decimal: for example: 1101 => 1*2^0 + 0*2^1 + 1*2^2 + 1*2^3 */ result += item * i32::pow(2, size); } result } }