[leetcode]445. Add Two Numbers II

Analysis

ummmm—— [每天刷题并不难。。。]

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.

因为两个数相加应该从低位加起，但是题目提到最好不要改动链表，所以我们可以把链表中的数存到stack中，然后再进行计算。

Implement

/**
 * 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> num1;
        stack<int> num2;
        while(l1){
            num1.push(l1->val);
            l1 = l1->next;
        }
        while(l2){
            num2.push(l2->val);
            l2 = l2->next;
        }
        int sum = 0;
        //int tmp = 0;
        ListNode* res = new ListNode(0);
        while(!num1.empty() || !num2.empty()){
            if(!num1.empty()){
                sum += num1.top();
                num1.pop();
            }
            if(!num2.empty()){
                sum += num2.top();
                num2.pop();
            }
            res->val = sum % 10;
            ListNode* tmp = new ListNode(sum/10);
            tmp->next = res;
            res = tmp;
            sum /= 10;
        }
        if(res->val == 0)
            return res->next;
        else
            return res;
    }
};