/**
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
}
}
