/**
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
    }
}