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.

Code

一開始寫了個又臭又長的版本。

/**
 * 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) {
        int carry = 0;
        ListNode dummy(0);
        ListNode* curr = &dummy;
        while(l1 && l2) {
            int val = (l1->val + l2->val + carry) % 10;
            curr->next = new ListNode(val);
            carry = (l1->val + l2->val + carry) / 10;
            curr = curr->next;
            l1 = l1->next;
            l2 = l2->next;
        }
 
        while(l1) {
            int val = (l1->val + carry) % 10;
            curr->next = new ListNode(val);
            carry = (l1->val + carry) / 10;
            curr = curr->next;
            l1 = l1->next;
        }
 
        while(l2) {
            int val = (l2->val + carry) % 10;
            curr->next = new ListNode(val);
            carry = (l2->val + carry) / 10;
            curr = curr->next;
            l2 = l2->next;
        }
 
        if(carry) {
            curr->next = new ListNode(carry);
        }
        
        return dummy.next;
    }
};

while 迴圈改用 or 就可以將 code 變簡潔。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
   int carry = 0;
   struct ListNode* dummy =  (struct ListNode*) malloc(sizeof(struct ListNode));
   dummy->next = NULL;
   struct ListNode* cur = dummy;
    while(l1 || l2 || carry) {
        int sum = 0;
        if(l1) {
            sum += l1->val;
            l1 = l1->next;
        }
            
        if(l2) {
            sum += l2->val;
            l2 = l2->next;
        }
 
        sum += carry;
        carry = sum / 10;
 
        struct ListNode* new_node =  (struct ListNode*) malloc(sizeof(struct ListNode));
        new_node->val = sum % 10;
        new_node->next = NULL;
 
        cur->next = new_node;
        cur = cur->next;
 
    }
   return dummy->next; 
}

/**
 * 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) {
        int carry = 0;
        ListNode dummy(0);
        ListNode* curr = &dummy;
        while(l1 || l2 || carry) {
            int sum = 0;
            if(l1) {
                sum += l1->val;
                l1 = l1->next;
            }
 
            if(l2) {
                sum += l2->val;
                l2 = l2->next;
            }
            sum += carry;
 
            curr->next = new ListNode(sum % 10);
            carry = sum / 10;
            curr = curr->next;
        }
        
        return dummy.next;
    }
};

Source

• Add Two Numbers - LeetCode