Add Two Numbers--LeetCode进阶路②

• 题目描述：

  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 contain a single digit. Add the two numbers and return it as a linked list.

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

  Example:

  Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
  Output: 7 -> 0 -> 8
  Explanation: 342 + 465 = 807.
  

• 思路分析：依旧温柔风的题儿~建个新链表，把两个链表求和撸一遍，需要注意求和时的进位问题
• 代码实现：

  /**
   * Definition for singly-linked list.
   * public class ListNode {
   *     int val;
   *     ListNode next;
   *     ListNode(int x) { val = x; }
   * }
   */
  class Solution {
      public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
          if(l1 == null || l2 == null){
              return l1 != null ? l1:l2;
          }
  
          int sum = l1.val + l2.val;
          ListNode result = new ListNode(sum % 10);
          result.next = addTwoNumbers(l1.next,l2.next);
  
          if(sum >= 10){//需要进位的情况
              result.next = addTwoNumbers(result.next,new ListNode(sum / 10));
          }
  
          return result;
      }
  }

  #现在的方法相当粗暴，现在给自己的目标是每天一题（并没有什么质量的那种啊啊啊），小陌考完四级要好好想想如何优化刷的题