2019年7月20日 - 2

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

我的方法:

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
11         ListNode ans = new ListNode(0);
12         ans.next = null;
13         ListNode curr = ans;
14
15         int c = 0;
16         while(l1.next != null && l2.next != null){
17             curr.val = (l1.val + l2.val + c)%10;
18             c = (l1.val + l2.val + c)/10;
19             curr.next = new ListNode(0);
20             curr = curr.next;
21             l1 = l1.next;
22             l2 = l2.next;
23         }
24         //两数等长度时同时处理最后一个数即可
25         if(l1.next == null && l2.next == null){
26             curr.val = (l1.val + l2.val + c)%10;
27             c = (l1.val + l2.val +c)/10;
28             //看最终是否还有一位进位
29             if(c == 1){
30                 curr.next = new ListNode(c);
31                 curr.next.next = null;
32             }else{
33                 curr.next = null;
34             }
35             return ans;
36         }
37         //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
38         ListNode l;
39         if(l1.next != null && l2.next == null){
40             l = l1;
41             curr.val = l2.val;
42         }else{
43             l = l2;
44             curr.val = l1.val;
45         }
46         //处理短的的最后一位
47         curr.val += l.val + c;
48         c = curr.val / 10;
49         curr.val %= 10;
50         curr.next = new ListNode(0);
51         curr = curr.next;
52         l = l.next;
53
54         while(l.next != null){
55             curr.val = (l.val + c)%10;
56             c = (l.val + c)/10;
57             curr.next = new ListNode(0);
58             curr = curr.next;
59             l = l.next;
60         }
61
62         //处理长的的最后一位
63         curr.val = (l.val + c)%10;
64         c = (l.val + c)/10;
65         if(c == 1){
66             curr.next = new ListNode(c);
67             curr.next.next = null;
68         }else{
69             curr.next = null;
70         }
71
72         return ans;
73     }
74 }

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)

原文地址：https://www.cnblogs.com/IsuxizWorld/p/11219194.html