leetcode-445. Add Two Numbers II

欢迎访问我的leetcode题解目录https://blog.csdn.net/richenyunqi/article/details/80689211

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

算法设计：

遍历两个给定链表，将两个链表中的元素分别放入两个栈中，然后两个栈同时弹出元素，进行加和，用头插法插入新链表中。

C++代码：

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        stack<int>s1,s2;
        for(;l1!=nullptr;l1=l1->next)
            s1.push(l1->val);
        for(;l2!=nullptr;l2=l2->next)
            s2.push(l2->val);
        int carry=0;//进位
        ListNode*l3=nullptr;//新链表头指针
        while(!s1.empty()||!s2.empty()||carry!=0){
            int k1=0,k2=0;
            if(!s1.empty()){
                k1=s1.top();
                s1.pop();
            }
            if(!s2.empty()){
                k2=s2.top();
                s2.pop();
            }
            ListNode*p=new ListNode((k1+k2+carry)%10);//新建结点
            p->next=l3;
            l3=p;//更新头指针
            carry=(k1+k2+carry)/10;//更新进位
        }
        return l3;
    }
};