2. Add Two Numbers && 371. Sum of Two Integers && 43. Multiply Strings

2. Add Two Numbers

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

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 Linked List Math

 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        
        if(l1 == null && l2 == null)
            return null;
        if(l2 == null)
            return l1;
        if(l1 == null)
            return l2;
        
        int previous = 0;
        int sum = l1.val+l2.val;
        previous = sum/10;
        ListNode r = new ListNode(sum%10);
        ListNode rc = r;
        
        l1 = l1.next;
        l2 = l2.next;
        while(l1 != null && l2 != null)
        {
            sum = l1.val+l2.val + previous;
            previous = sum/10;
            r.next = new ListNode(sum%10);
            r = r.next;
            l1 = l1.next;
            l2 = l2.next;
        }
        
        while(l1 != null)
        {
            sum = l1.val + previous;
            previous = sum/10;
            r.next = new ListNode(sum%10);
            r = r.next;
            l1 = l1.next;
        }
        
        while(l2 != null)
        {
            sum = l2.val + previous;
            previous = sum/10;
            r.next = new ListNode(sum%10);
            r = r.next;
            l2 = l2.next;
        }
        
        if(previous != 0)
            r.next = new ListNode(1);
        return rc;
        
    }
}

 

371. Sum of Two Integers

Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.

 Bit Manipulation

 

Solution:

For example, we have a = 1, b = 3. In bit representation, a = 0001, b = 0011. First, we can use "and"("&") operation between a and b to find a carry.

carry = a & b, then carry = 0001

Second, we can use "xor" ("^") operation between a and b to find the different bit, and assign it to a. Then, we shift carry one position left and assign it to b, b = 0010.

Iterate until there is no carry (or b == 0)

 

public class Solution {
    public int getSum(int a, int b) {
        if (a == 0) return b;
        if (b == 0) return a;
    
        while (b != 0) {
            int carry = a & b;
            a = a ^ b;
            b = carry << 1;
        }
        
        return a;
    }
}

 

43. Multiply Strings

Given two numbers represented as strings, return multiplication of the numbers as a string.

Note:

• The numbers can be arbitrarily large and are non-negative.
• Converting the input string to integer is NOT allowed.
• You should NOT use internal library such as BigInteger.

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 Math String

 

public class Solution {
  public String multiply(String num1, String num2) {
    int l1 = num1.length();
    int l2 = num2.length();
    int[] product = new int[l1 + l2];

    for (int i = l1 - 1; i >= 0; --i) {
      for (int j = l2 - 1; j >= 0; --j) {
        int posTens = i + j;
        int posOnes = posTens + 1;

        int sum = product[posOnes] + (num1.charAt(i) - '0') * (num2.charAt(j) - '0');

        product[posTens] += sum / 10;
        product[posOnes] = (sum) % 10;
      }
    }

    StringBuilder sb = new StringBuilder();
    for (int num : product)
      if (sb.length() > 0 || num > 0)
        sb.append(num);

    return sb.length() == 0 ? "0" : sb.toString();
  }
}