【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。

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        stack<int> s1, s2;
        while (l1) {
            s1.push(l1->val);
            l1 = l1->next;
        }
        while (l2) {
            s2.push(l2->val);
            l2 = l2->next;
        }
        int sum = 0;
        ListNode* pre = nullptr;
        while (!s1.empty() || !s2.empty() || sum) {
            if (!s1.empty()) {
                sum += s1.top(); 
                s1.pop();
            }
            if (!s2.empty()) {
                sum += s2.top(); 
                s2.pop();
            }
            ListNode *node = new ListNode(sum % 10);
            node->next = pre;
            pre = node;
            sum /= 10;
        }
        return pre;
    }
};

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。

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int len1 = getLength(l1), len2 = getLength(l2);
        if (len1 < len2) 
            swap(l1, l2);
        ListNode* dummy = new ListNode(0);
        ListNode* cur = dummy, *right = dummy;
        int diff = abs(len1 - len2);
        
        while (diff) {
            cur->next = new ListNode(l1->val);
            cur = cur->next;
            if (cur->val != 9)
                right = cur;
            l1 = l1->next;
            --diff;
        }
        while (l1) {
            int sum = l1->val + l2->val;
            cur->next = new ListNode(sum % 10);
            cur = cur->next;
            if (sum > 9) {
                sum %= 10;
                ++right->val;
                right = right->next;
                while (right!= cur) {
                    right->val = 0;
                    right = right->next;
                }
            } 
            if (sum != 9)
                right = cur;
            l1 = l1->next;
            l2 = l2->next;
        }
        
        return dummy->val == 0 ? dummy->next : dummy;
    }
    int getLength(ListNode* head) {
        if (!head)
            return 0;
        int len = 0;
        while (head) {
            ++len;
            head = head->next;
        }
        return len;
    }
};