[445] 两数相加 II

给你两个 非空 链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。将这两数相加会返回一个新的链表。
你可以假设除了数字 0 之外，这两个数字都不会以零开头。

进阶：
如果输入链表不能修改该如何处理？换句话说，你不能对列表中的节点进行翻转。

示例：

输入：(7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 8 -> 0 -> 7

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        stack<int> h_1;
        stack<int> h_2;
        while(l1 != NULL){
            h_1.push(l1->val);
            l1 = l1->next;
        }
        while(l2 != NULL){
            h_2.push(l2->val);
            l2 = l2->next;
        }
        int length = max(h_1.size(), h_2.size());
        ListNode* res = new ListNode(0);
        ListNode* head = res;
        int flag = 0;

        while(!h_1.empty() || !h_2.empty()){
            int value;
            if(!h_1.empty() && !h_2.empty()){
                value = (h_1.top() + h_2.top() + flag) % 10;
                flag = (h_1.top() + h_2.top() + flag) / 10;
                h_1.pop();
                h_2.pop();
            }else if(!h_1.empty()){
                value = (h_1.top() + flag) % 10;
                flag = (h_1.top() + flag) / 10;   
                h_1.pop();             
            }else{
                value = (h_2.top() + flag) % 10;
                flag = (h_2.top() + flag) / 10;   
                h_2.pop();             
            }
            ListNode* next = new ListNode(value);
            res->next = next;
            res = res->next; 
        }
        if(flag > 0){
            ListNode* next = new ListNode(flag);
            res->next = next;
            res = res->next; 
        }
        ListNode* pre = NULL;
        ListNode* cur = head->next;
        while(cur != NULL){
            ListNode* next = cur->next;
            cur->next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
};

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 carry = 0;
        ListNode* ans = nullptr;
        while (!s1.empty() or !s2.empty() or carry != 0) {
            int a = s1.empty() ? 0 : s1.top();
            int b = s2.empty() ? 0 : s2.top();
            if (!s1.empty()) s1.pop();
            if (!s2.empty()) s2.pop();
            int cur = a + b + carry;
            carry = cur / 10;
            cur %= 10;
            auto curnode = new ListNode(cur);
            curnode -> next = ans;
            ans = curnode;
        }
        return ans;
    }
};

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/add-two-numbers-ii/solution/liang-shu-xiang-jia-ii-by-leetcode-solution/
来源：力扣（LeetCode）
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