Add Two Numbers II

Add Two Numbers II

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

解析： 开始以为数据在int范围内后来意识到时不限位数只能像字符串相加一样计算

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) {
        vector<int>num1;
        vector<int>num2;
        if (l1==NULL||l2==NULL) return NULL;
        while(l1)
        {
           num1.push_back(l1->val);
           l1=l1->next;
        }
        while(l2)
        {
            num2.push_back(l2->val);
            l2=l2->next;
        }
       
        int bit=0;
        int p1=num1.size();
        int p2=num2.size();
         vector<int>ans;
        // int curbit=ans.size()-1;
        while(p1&&p2)
        {
            int bitsum=num1[p1-1]+num2[p2-1];
            bitsum+=bit;
            ans.push_back(bitsum%10);
            bit=bitsum/10;
            p1--;
            p2--;
            
        }
        
        while(p1)
        {
            int bitsum=num1[p1-1]+bit;
            ans.push_back(bitsum%10);
            bit=bitsum/10;
            p1--;
        }
          while(p2)
        {
            int bitsum=num2[p2-1]+bit;
            ans.push_back(bitsum%10);
            bit=bitsum/10;
            p2--;
        }
        if (bit)
        {
            ans.push_back(bit);
        }
        
        ListNode * pnode;
        if (!ans.empty())
        pnode=new ListNode(ans[ans.size()-1]);
 
      
        ListNode * anshead=pnode;
        for (int i=ans.size()-2; i>=0; i--)
        {
            pnode->next=new ListNode(ans[i]);
            pnode=pnode->next;
        }
        pnode->next=NULL;
        
        return anshead;
        
        
        
        
    }
};