【LeetCode】445. Add Two Numbers II

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

题解：

Solution 1

　　加和是从尾部开始的，而链表又只能从前开始遍历，那么这种单向的遍历顺序就及其适合用queue或者stack来辅助。显然此题用到的是stack。

 1 class Solution {
 2 public:
 3     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
 4         stack<int> s1, s2;
 5         while (l1) {
 6             s1.push(l1->val);
 7             l1 = l1->next;
 8         }
 9         while (l2) {
10             s2.push(l2->val);
11             l2 = l2->next;
12         }
13         int sum = 0;
14         ListNode* pre = nullptr;
15         while (!s1.empty() || !s2.empty() || sum) {
16             if (!s1.empty()) {
17                 sum += s1.top(); 
18                 s1.pop();
19             }
20             if (!s2.empty()) {
21                 sum += s2.top(); 
22                 s2.pop();
23             }
24             ListNode *node = new ListNode(sum % 10);
25             node->next = pre;
26             pre = node;
27             sum /= 10;
28         }
29         return pre;
30     }
31 };

Solution 2

　　设计的很巧妙，无需开辟存储空间。转自 Grandyang。用right记录目前最右第一个值不是9的元素。用cur遍历链表。如果cur遍历时有进位，那么right自增1，且right到cur（左开右闭）之间的由于都是9（因为right指向的是目前最右第一个值不是9的元素），所以此区间内的元素值全部置0。

 1 class Solution {
 2 public:
 3     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
 4         int len1 = getLength(l1), len2 = getLength(l2);
 5         if (len1 < len2) 
 6             swap(l1, l2);
 7         ListNode* dummy = new ListNode(0);
 8         ListNode* cur = dummy, *right = dummy;
 9         int diff = abs(len1 - len2);
10         
11         while (diff) {
12             cur->next = new ListNode(l1->val);
13             cur = cur->next;
14             if (cur->val != 9)
15                 right = cur;
16             l1 = l1->next;
17             --diff;
18         }
19         while (l1) {
20             int sum = l1->val + l2->val;
21             cur->next = new ListNode(sum % 10);
22             cur = cur->next;
23             if (sum > 9) {
24                 sum %= 10;
25                 ++right->val;
26                 right = right->next;
27                 while (right!= cur) {
28                     right->val = 0;
29                     right = right->next;
30                 }
31             } 
32             if (sum != 9)
33                 right = cur;
34             l1 = l1->next;
35             l2 = l2->next;
36         }
37         
38         return dummy->val == 0 ? dummy->next : dummy;
39     }
40     int getLength(ListNode* head) {
41         if (!head)
42             return 0;
43         int len = 0;
44         while (head) {
45             ++len;
46             head = head->next;
47         }
48         return len;
49     }
50 };