1290. Convert Binary Number in a Linked List to Integer.cpp

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:

The Linked List is not empty.
Number of nodes will not exceed 30.
Each node's value is either 0 or 1.
   Hide Hint #1
Traverse the linked list and store all values in a string or array. convert the values obtained to decimal value.
   Hide Hint #2
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.








/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    int getDecimalValue(ListNode* head) {
        int len=0;
        ListNode *curr=head;
        while(curr){
            len++;
            curr=curr->next;
        }
        curr=head;
        len--;
        int res=0;

        while(curr){
            res+=curr->val*pow(2, len);
            len--;
            curr=curr->next;
        }
        return res;
    }
};