2019年7月20日 - 2

 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 }