使用Java中的单个链表和递归方法将二进制转换为十进制

Question

How we can convert binary to decimal using single linked list and recursive method in java? :-s

Ex:

Input: 1->0->0->NULL

Output: 4

Answer 1

It's a bit tricky recursively because you need to scale "earlier" values depending on the list length, simplest way seems to be to hand down the result "so far" via an additional parameter.

class Node {
  int value;
  Node next;

  int toDecimal(int resultSoFar) {
    int resultSoFar = 2 * resultSoFar + value;
    return next == null ? resultSoFar : toDecimal(resultSoFar);
  } 

  int toDecimal() {
    toDecimal(0);
  }
}

Answer 2

Try this (Without the null at the end of the list) :

public Integer binToDec(LinkedList<Integer> list){
        Double size = (double) list.size();
        if(size == 0D) return 0;
        Integer number = list.getFirst();
        list.removeFirst();
        if(number != 0) return binToDec(list) + number*new Double(Math.pow(2D, size - 1)).intValue();
        else return binToDec(list);
    }

Be aware that it will clear the list.

Answer 3

I can think of two ways to solve it:

1- If length of list is known:

// To find length of list
public int length(Node head) {
      int count = 0;
      while(head != null) {
         count++;
         head = head.next;
      }
      return count;
   }

public static int convertWhenLengthIsKnown(Node head, int len) {
      int sum = 0;
      if(head.next != null) {
         sum = convertWhenLengthIsKnown(head.next, len-1);
      }
      return sum + head.data * (int)Math.pow(2,len);
   }


   // Call this function as below:

convertWhenLengthIsKnown(head, length(head)-1);

2. If we don't want to calculate length, then we can have a sum variable which is globally accessible,

    private static int sum = 0; public static int convert(Node head,int i) { if(head.next != null) { i = convert(head.next, i); } sum += head.data * (int)Math.pow(2,i); return i+1; } // Call this function as below: convert(head,0); 

Below is the Node class:

class Node {
   int data;
   Node next;

   Node(int data) {
      this.data = data;
   }
}

Hope It helps you.

Answer 4

As you did not tell anything about your linked list, I assume it's similar to the Node class from pbajpai21 .

Without knowing the length of the chain you could with each level shift the value one position to the left.

the class for each node in the list

class Node {
    int digit;
    Node child;

    Node(int data) {
        this.digit = data;
    }

    public int getDigit() {
        return digit;
    }

    public void setChild(Node child) {
        this.child = child;
    }

    public Node getChild() {
        return child;
    }
}

the class for demonstration

public class Bits {
    public static void main(String[] args) {
        int[] ints = new int[] {1, 0, 1, 0, 1, 0};
        Node parent = new Node(ints[0]);
        Node root = parent;
        for (int i = 1; i < ints.length; i++) {
            Node child = new Node(ints[i]);
            parent.setChild(child);
            parent = child;
        }

        long value = computeValue(0, root);
        System.out.println();
        System.out.println("value = " + value);
    }

    private static long computeValue(long parentValue, Node node) {
        if (node == null) {
            return parentValue;
        }
        // only to print the current digit
        System.out.print(node.getDigit());
        long value = (parentValue << 1) + node.getDigit();
        return computeValue(value, node.getChild());
    }
}

output

101010
value = 42