Add Two Numbers

https://leetcode.com/problems/add-two-numbers/description/

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:

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.

# Python
# Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution:
    def addTwoNumbers(self, l1,l2):
        ans = ListNode()
        temp = ans
        carry = 0

        while l1 or l2 or carry !=0 :
            if l1:
                d1 = l1.val
            else : d1 = 0

            if l2:
                d2 = l2.val
            else : d2 = 0

            sum = d1 + d2 + carry
            d = sum % 10
            carry = sum // 10

            newNode = ListNode(d)
            temp.next = newNode
            temp = temp.next

            if l1:
                l1 = l1.next
            if l2:
                l2 = l2.next

        return ans.next

// C언어
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode* createNode(int val) {
    struct ListNode* newNode = (struct ListNode*)malloc(sizeof(struct ListNode));
    newNode->val = val;
    newNode->next = NULL;
    return newNode;
}

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode ans;
    struct ListNode* temp = &ans;
    int c = 0;
    int d1,d2;
    while(l1 || l2 || c!=0 ){
        if(l1){
            d1 = l1->val;
        }else d1 = 0;
        if(l2){
            d2 = l2->val;
        }else d2 = 0;

        int sum = (d1+d2+c)%10;
        c = (d1+d2+c)/10;

        struct ListNode *newNode = createNode(sum);

        temp -> next = newNode;
        temp = temp->next;

        if(l1){
            l1 = l1->next;
        }
        if(l2){
            l2 = l2->next;
        }
    }
    return ans.next;

}

 

C로 했을 땐 좀 귀찮게

 

struct Listnode* createNode(int val){

 

struct Listnode* newNode = (struct Listnode*)malloc(sizeof(struct Listnode));

newNode -> val = val;

newNode -> next = NULL;

 

return newNode;

};

 

이렇게 귀찮지만 함수를 선언해주는 편이 좋다.

 

//C++ 코드
/**
 * 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* ans = new ListNode(999);
        ListNode* temp = ans;
        int c = 0;

        while(l1||l2||c!=0){
            int d1 = (l1) ? l1->val : 0;
            int d2 = (l2) ? l2->val : 0;

            int sum = (d1+d2+c);
            int d = sum % 10;
            c = sum/10;

            ListNode* newNode = new ListNode(d);
            temp -> next = newNode;
            temp = temp -> next;

            l1 = (l1) ? l1->next : nullptr;
            l2 = (l2) ? l2->next : nullptr;
        }
        ListNode* result = ans->next;
        delete ans;
        return result;
    }
};