Two Sum: Input array is sorted

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Two Sum: Input array is sorted

Similar to Question [1. Two Sum], except that the input array is already sorted in ascending order.

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

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2

Input: l1 = [0], l2 = [0]
Output: [0]

Example 3

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

Farjanul Answered question July 27, 2022
0

Of course we could still apply the [Hash table] approach, but it costs us O(n) extra space, plus it does not make use of the fact that the input is already sorted.

O(n log n) runtime, O(1) space – Binary search:
For each element x, we could look up if target – x exists in O(log n) time by applying binary search over the sorted array. Total runtime complexity is O(n log n).

Solution in Java

public class Solution {
    // example in leetcode book
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
     ListNode dummyHead = new ListNode(0);
     ListNode p = l1, q= l2, curr = dummyHead;
     int carry = 0;
     while (p != null || q!= null) {
      int x = (p != null) ? p.val : 0;
      int y = (q != null) ? q.val : 0;
      int digit = carry + x + y;
      carry = digit / 10;
      curr.next = new ListNode(digit % 10);
      curr = curr.next;
      if (p != null) p = p.next;
      if (q != null) q = q.next;
     }
     if (carry > 0) {
      curr.next = new ListNode(carry);
     }
     return dummyHead.next;
    }
}

Solution in Python

class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        carry = 0
        # dummy head
        head = curr = ListNode(0)
        while l1 or l2:
            val = carry
            if l1:
                val += l1.val
                l1 = l1.next
            if l2:
                val += l2.val
                l2 = l2.next
            curr.next = ListNode(val % 10)
            curr = curr.next
            carry = int(val / 10)
        if carry > 0:
            curr.next = ListNode(carry)
        return head.next

Farjanul Edited answer July 27, 2022
0