2019年7月20日 - LeetCode0002

https://leetcode-cn.com/problems/add-two-numbers/submissions/

我的方法:

 /**
  * 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) {
         ListNode ans = new ListNode(0);
         ans.next = null;
         ListNode curr = ans;

         int c = 0;
         while(l1.next != null && l2.next != null){
             curr.val = (l1.val + l2.val + c)%10;
             c = (l1.val + l2.val + c)/10;
             curr.next = new ListNode(0);
             curr = curr.next;
             l1 = l1.next;
             l2 = l2.next;
         }
         //两数等长度时同时处理最后一个数即可
         if(l1.next == null && l2.next == null){
             curr.val = (l1.val + l2.val + c)%10;
             c = (l1.val + l2.val +c)/10;
             //看最终是否还有一位进位
             if(c == 1){
                 curr.next = new ListNode(c);
                 curr.next.next = null;
             }else{
                 curr.next = null;
             }
             return ans;
         }
         //一长一短时,留下长的那个,把短的的最后一个val记录后用不着了
         ListNode l;
         if(l1.next != null && l2.next == null){
             l = l1;
             curr.val = l2.val;
         }else{
             l = l2;
             curr.val = l1.val;
         }
         //处理短的的最后一位
         curr.val += l.val + c;
         c = curr.val / 10;
         curr.val %= 10;
         curr.next = new ListNode(0);
         curr = curr.next;
         l = l.next;

         while(l.next != null){
             curr.val = (l.val + c)%10;
             c = (l.val + c)/10;
             curr.next = new ListNode(0);
             curr = curr.next;
             l = l.next;
         }

         //处理长的的最后一位
         curr.val = (l.val + c)%10;
         c = (l.val + c)/10;
         if(c == 1){
             curr.next = new ListNode(c);
             curr.next.next = null;
         }else{
             curr.next = null;
         }

         return ans;
     }
 }

按部就班的考虑与处理.注意不要有考察漏掉的情况

时间复杂度O(n),空间复杂度O(n)

//我所有的时空复杂度都是指级别,有必要具体分析的会特别注明

结果:

官方题解方法:

https://leetcode-cn.com/problems/add-two-numbers/solution/liang-shu-xiang-jia-by-leetcode/

也是朴素方法,区别在于他没有拆开来考虑,而是用了或判断,在末尾前进的时候又加了if

代码量比我的少,优雅一些.

(见仁见智,我不喜欢在简单的循环里加判断,徒增复杂度)