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 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.
Example:
- Input:
2 -> 4 -> 3+5 -> 6 -> 4 - Output:
7 -> 0 -> 8
C Solution:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
struct ListNode dummy, *tail = &dummy;
int carry = 0;
while (l1 || l2 || carry) {
if (l1) {
carry += l1->val;
l1 = l1->next;
}
if (l2) {
carry += l2->val;
l2 = l2->next;
}
tail->next = malloc(sizeof(struct ListNode));
tail->next->val = carry % 10;
tail = tail->next;
carry /= 10;
}
tail->next = 0;
return dummy.next;
}
Python Solution:
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
head = dummy = ListNode(0)
carry = 0
while l1 or l2 or carry:
if l1:
carry += l1.val
l1 = l1.next
if l2:
carry += l2.val
l2 = l2.next
head.next = head = ListNode(carry % 10)
carry //= 10
return dummy.next
Summary:
- 19ms, 39.69%
- I need to remember the new list's head and I need to track the list's tail, then a dummy head like this is perfect.
LeetCode: 2. Add Two Numbers
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