Given a singly linked list of 0s and 1s find its decimal equivalent.
Input: 0->0->0->1->1->0->0->1->0 Output: 50 Input: 1->0->0 Output: 4
The decimal value of an empty linked list is considered as 0.
Initialize the result as 0. Traverse the linked list and for each node, multiply the result by 2 and add the node's data to it.
// C++ Program to find decimal value
// of binary linked list
#include <bits/stdc++.h>
using namespace std;
// Link list Node
class Node
{
public:
bool data;
Node* next;
};
/* Returns decimal value of binary
linked list */
int decimalValue(Node *head)
{
// Initialized result
int res = 0;
// Traverse linked list
while (head != NULL)
{
// Multiply result by 2 and
// add head's data
res = (res << 1) + head->data;
// Move next
head = head->next;
}
return res;
}
// Utility function to create a
// new node.
Node *newNode(bool data)
{
Node *temp = new Node;
temp->data = data;
temp->next = NULL;
return temp;
}
// Driver code
int main()
{
// Start with the empty list
Node* head = newNode(1);
head->next = newNode(0);
head->next->next = newNode(1);
head->next->next->next = newNode(1);
cout << "Decimal value is " <<
decimalValue(head);
return 0;
}
// This is code is contributed by rathbhupendra
Output :
Decimal value is 11
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Another Approach(by reversing the Linked List):
Follow the below steps to solve the given problem
1) First reverse the given linked list.
2) Initialize a ans variable to store ans and pos variable to keep track of position of node in linked list.
3) Perform the operation ans = ans + (rhead.data*(2**pos))%MOD)
4) perform ans = ans%MOD
Below is the implementation of above approach:
#include<bits/stdc++.h>
using namespace std;
// C++ Program to find decimal value
// of binary linked list
// node structure
struct Node{
int data;
Node* next;
Node(int data){
this->data = data;
this->next = NULL;
}
};
long long unsigned int power(int num,int count){
if(count ==0) return 1;
if(count%2==0)
return (power(num,count/2)%1000000007)*(power(num,count/2)%1000000007);
else
return num*(power(num,count/2)%1000000007)*(power(num,count/2)%1000000007);;
}
Node* reverse(Node* head){
if(head == NULL || head->next == NULL) return head;
Node* curr = head;
Node* prev = NULL;
Node* nxt = head->next;
while(nxt != NULL){
curr->next = prev;
prev = curr;
curr = nxt;
nxt = nxt->next;
}
curr->next = prev;
return curr;
}
int decimalValue(Node* head){
int MOD = 1000000007;
Node* rhead = reverse(head);
int ans = 0;
int pos = 0;
while(rhead != NULL){
ans = (ans%MOD + ((rhead->data)*power(2,pos)) % MOD) % MOD;
rhead = rhead->next;
pos++;
}
return ans;
}
int main(){
Node* head = new Node(1);
head->next = new Node(0);
head->next->next = new Node(1);
head->next->next->next = new Node(1);
cout<<"Decimal Value is : "<<decimalValue(head);
}
Output
Decimal Value is : 11
Time Complexity: O(N) where N is the number of nodes in linked list.
Auxiliary Space: O(1)
Please refer complete article on Decimal Equivalent of Binary Linked List for more details!
