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Add Two Numbers.cpp
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/*
Add Two Numbers
===============
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.
*/
/**
* 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 *dummy = new ListNode(-1);
auto temp = dummy;
int carry = 0;
while (l1 && l2)
{
int digit = l1->val + l2->val + carry;
carry = digit / 10;
digit = digit % 10;
l1 = l1->next;
l2 = l2->next;
ListNode *newNode = new ListNode(digit);
temp->next = newNode;
temp = temp->next;
}
while (l1)
{
int digit = l1->val + carry;
carry = digit / 10;
digit = digit % 10;
l1 = l1->next;
ListNode *newNode = new ListNode(digit);
temp->next = newNode;
temp = temp->next;
}
while (l2)
{
int digit = l2->val + carry;
carry = digit / 10;
digit = digit % 10;
l2 = l2->next;
ListNode *newNode = new ListNode(digit);
temp->next = newNode;
temp = temp->next;
}
while (carry)
{
int digit = carry;
carry = digit / 10;
digit = digit % 10;
ListNode *newNode = new ListNode(digit);
temp->next = newNode;
temp = temp->next;
}
return dummy->next;
}
};