版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/w5688414/article/details/86557501
Description
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
分析
- 这里用两个栈来模拟加法。
- 代码我觉得写得直观易懂，直接看代码吧
代码
/**
* 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 sum=0;
ListNode *res=new ListNode(sum);
while(!s1.empty()||!s2.empty()){
if(!s1.empty()){
sum+=s1.top();s1.pop();
}
if(!s2.empty()){
sum+=s2.top();s2.pop();
}
res->val=sum%10;
ListNode* head=new ListNode(sum/10);
head->next=res;
res=head;
sum/=10;
}
return res->val==0 ? res->next:res;
}
};