0002.cpp

// Source: https://leetcode.com/problems/add-two-numbers
// Title: Add Two Numbers
// Difficulty: Medium
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
//
// You may assume that each input would have exactly one solution, and you may not use the same element twice.
//
// You can return the answer in any order.
//
// Example 1:
//
//   Input: nums = [2,7,11,15], target = 9
//   Output: [0,1]
//   Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
//
// Example 2:
//
//   Input: nums = [3,2,4], target = 6
//   Output: [1,2]
//
// Example 3:
//
//   Input: nums = [3,3], target = 6
//   Output: [0,1]
//
// Constraints:
//
//   2 <= nums.length <= 10^4
//   -10^9 <= nums[i] <= 10^9
//   -10^9 <= target <= 10^9
//   Only one valid answer exists.
//
// Follow-up: Can you come up with an algorithm that is less than O(n^2) time complexity?
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

class Solution {
 public:
  ListNode* addTwoNumbers( ListNode* l1, ListNode* l2, int carry = 0 ) {
    if ( !l1 && !l2 ) {
      return carry ? new ListNode(carry) : nullptr;
    }
    carry += (l1 ? l1->val : 0) + (l2 ? l2->val : 0);
    auto *l3 = new ListNode(carry%10);
    l1 = l1 ? l1->next : nullptr;
    l2 = l2 ? l2->next : nullptr;
    l3->next = addTwoNumbers(l1, l2, carry/10);
    return l3;
  }
};