Leetcode 1290 solution in java

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.

The problem is taken from the below mentioned link:- Leetcode Problem

Approach:-

The simplest approach is using Array List Data Structure.  To get the decimal value, the approach is :-

1*(2^2)+0*(2^1)+1*(2^0) = 4 + 0 + 1 = 5

So, store the value in an Array List Data Structure, as :-

1	0	1
Position 0	Position 1	Position 2

Then using for loop we can achieve our desired result as mentioned below:-

Int p = 0;

for(int i=list.size()-1;i>=0;i–){

            int pow = (1<<p);    //  Finding 2 to the power p

            res += (pow*list.get(i));

            p++;

        }

Flow Diagram:-

Code:-

	public int getDecimalValue(ListNode head) {

		int result = 0;
		ArrayList<Integer> list = new ArrayList<>();

		while (head != null) {
			list.add(head.val);
			head = head.next;
		}
		int p = 0;
		for (int i = list.size() - 1; i >= 0; i--) {
			int pow = (1 << p); // Finding 2 to the power p

			System.out.println("The value of pow is " + pow + " ant the size remaining  is " + i);

			result += (pow * list.get(i));
			p++;

			System.out.println("The value of the counter is " + p + " ant res value is " + result);
		}

		return result;
	}

For The entire code please follow the below section:-

package ArrayAll;

import java.util.ArrayList;

public class Solution {

	public static ListNode head = null;
	public static ListNode tail = null;

	public static int[] arr;

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Solution x = new Solution();

		x.insertDataAtEnd(1);

		x.insertDataAtEnd(0);
		x.insertDataAtEnd(1);
		x.display();

		System.out.println(x.getDecimalValue(head));

	}

	public int getDecimalValue(ListNode head) {

		int result = 0;
		ArrayList<Integer> list = new ArrayList<>();

		while (head != null) {
			list.add(head.val);
			head = head.next;
		}
		int p = 0;
		for (int i = list.size() - 1; i >= 0; i--) {
			int pow = (1 << p); // Finding 2 to the power p

			System.out.println("The value of pow is " + pow + " ant the size remaining  is " + i);

			result += (pow * list.get(i));
			p++;

			System.out.println("The value of the counter is " + p + " ant res value is " + result);
		}

		return result;
	}

	// method to add new node at the end

	public static void insertDataAtEnd(int data) {

		/// If only one node present

		ListNode newNode = new ListNode(data);

		if (head == null) {

			head = newNode;
			tail = newNode;
		}

		else {

			tail.next = newNode;
			tail = newNode;

		}
	}

	public static ListNode display() {
		int i = 0;
		ListNode current = head;

		if (head == null) {

			System.out.println("List is empty");
			return null;
		}

		while (current != null) {

			System.out.println(current.val + " ");

			current = current.next;

		}

		return current;

	}

}

class ListNode {
	int val;
	ListNode next;

	ListNode() {
	};

	ListNode(int val) {
		this.val = val;
	}

	ListNode(int val, ListNode next) {
		this.val = val;
		this.next = null;
	}
}

For more problem and solutions please visit the below mentioned link:- Algorithms and Data-structure/