2. Add Two Numbers

Question:
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:
2. Add Two Numbers
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.
Example 2:

Input: l1 = [0], l2 = [0]
Output: [0]
Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

Constraints:

The number of nodes in each linked list is in the range [1, 100].
0 <= Node.val <= 9
It is guaranteed that the list represents a number that does not have leading zeros.

Solution:

      /**
      * Definition for singly-linked list.
      * struct ListNode {
      *     int val;
      *     ListNode *next;
      *     ListNode() : val(0), next(nullptr) {}
      *     ListNode(int x) : val(x), next(nullptr) {}
      *     ListNode(int x, ListNode *next) : val(x), next(next) {}
      * };
      */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode *curcor = new ListNode(0);
        ListNode *temp = curcor;
        int carry = 0;
        while(l1 != NULL || l2 != NULL)
        {
            int num  = 0;
            if(l1 != NULL && l2 != NULL)
            {
             num = l1->val+l2->val;    
                l2 = l2->next;
                l1 = l1->next;
            }else if(l1 == NULL && l2 != NULL)
            {
                num = l2->val;    
                l2 = l2->next;
            }else if(l1 != NULL && l2 == NULL)
            {
                num = l1->val;    
                l1 = l1->next;
            }
            
            int current = (num+carry)%10;
            ListNode *p = new ListNode(current);
            temp->next =p;
            temp = p;
            carry = (num+carry)/10;
            if(l1 == NULL && l2 == NULL && carry != 0)
            {
            ListNode *p = new ListNode(carry);
            temp->next =p;
            temp = p;
            }
        }

        return curcor->next;
    }
};

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