链表－leetcode 445 Add Two Numbers

原题链接：Add two numbers II

题解：

//分析：这道题是Add two numbers的演化版，当然可以先将链表反转，然后回到了之前的那道题。但是题目明确follow not modify the input lists，就是禁止反转输入链表。怎么办呢？链表从高位到低位，加法从低位到高位，貌似这种关系可以借用到栈的先进后出嚎～。于是答案就来了：/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */#include <stack>class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        /*            Time Complexity:O(max(n,m))            Space Complexity:O(m+n)        */        if(!l1)return l2;        if(!l2)return l1;        stack<int>s1;        stack<int>s2;        while(l1){            s1.push(l1->val);            l1=l1->next;        }        while(l2){            s2.push(l2->val);            l2=l2->next;        }        int flag=0;        int temp=0;        ListNode* res=NULL;        while(!s1.empty() && !s2.empty()){            temp=s1.top()+s2.top()+flag;            s1.pop();            s2.pop();            flag=temp/10;            ListNode* ptr=new ListNode(temp%10);            ptr->next=res;            res=ptr;        }        while(!s1.empty()){            temp=s1.top()+flag;            s1.pop();            flag=temp/10;            ListNode* ptr=new ListNode(temp%10);            ptr->next=res;            res=ptr;        }        while(!s2.empty()){            temp=s2.top()+flag;            s2.pop();            flag=temp/10;            ListNode* ptr=new ListNode(temp%10);            ptr->next=res;            res=ptr;        }        if(flag){            ListNode* ptr=new ListNode(1);            ptr->next=res;            res=ptr;        }        return res;    }};