LEETCODE算法:445.Add Two Numbers II

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题目

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

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//栈，其实用string感觉更高效
/**
 * 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> s1,s2;
        while(l1)
        {
            s1.push(l1->val);
            l1=l1->next;
        }
        while(l2)
        {
            s2.push(l2->val);
            l2=l2->next;
        }
        int tmp=0;
        ListNode *res=NULL;
        while(!s1.empty()||!s2.empty()||tmp!=0)
        {
            int x1=0,x2=0;
            if(!s1.empty())
            {
                x1=s1.top();
                s1.pop();
            }
            if(!s2.empty())
            {
                x2=s2.top();
                s2.pop();
            }
            int sum=x1+x2+tmp;
            tmp=sum/10;
            sum%=10;
            auto node=new ListNode(sum);
            //node->val=sum;
            node->next=res;
            res=node;
        }
        return res;
    }
};