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给你两个 非空 的链表,表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的,并且每个节点只能存储 一位 数字。
请你将两个数相加,并以相同形式返回一个表示和的链表。
你可以假设除了数字 0 之外,这两个数都不会以 0 开头。
示例 1:
输入:l1 = [2,4,3], l2 = [5,6,4] 输出:[7,0,8] 解释:342 + 465 = 807.
示例 2:
输入:l1 = [0], l2 = [0] 输出:[0]
示例 3:
输入:l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9] 输出:[8,9,9,9,0,0,0,1]
提示:
[1, 100]
内0 <= Node.val <= 9
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
carry = 0
dummy = ListNode(-1)
cur = dummy
while l1 or l2 or carry:
s = (0 if not l1 else l1.val) + (0 if not l2 else l2.val) + carry
carry, val = divmod(s, 10)
cur.next = ListNode(val)
cur = cur.next
l1 = None if not l1 else l1.next
l2 = None if not l2 else l2.next
return dummy.next
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int carry = 0;
ListNode dummy = new ListNode(-1);
ListNode cur = dummy;
while (l1 != null || l2 != null || carry != 0) {
int s = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
carry = s / 10;
cur.next = new ListNode(s % 10);
cur = cur.next;
l1 = l1 == null ? null : l1.next;
l2 = l2 == null ? null : l2.next;
}
return dummy.next;
}
}
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
int carry = 0;
ListNode* dummy = new ListNode(-1);
ListNode* cur = dummy;
while (l1 != NULL || l2 != NULL || carry != 0) {
int s = (l1 == NULL ? 0 : l1-> val) + (l2 == NULL ? 0 : l2->val) + carry;
carry = s / 10;
cur->next = new ListNode(s % 10);
cur = cur->next;
l1 = l1 == NULL ? NULL : l1->next;
l2 = l2 == NULL ? NULL : l2->next;
}
return dummy->next;
}
};
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function (l1, l2) {
let carry = 0;
const dummy = new ListNode(-1);
let cur = dummy;
while (l1 || l2 || carry) {
const s = (l1 ? l1.val : 0) + (l2 ? l2.val : 0) + carry;
carry = Math.floor(s / 10);
cur.next = new ListNode(s % 10);
cur = cur.next;
l1 = l1 ? l1.next : l1;
l2 = l2 ? l2.next : l2;
}
return dummy.next;
};
/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int val=0, ListNode next=null) {
* this.val = val;
* this.next = next;
* }
* }
*/
public class Solution {
public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(-1);
ListNode cur = dummy;
var carry = 0;
while (l1 != null || l2 != null || carry != 0)
{
int t = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
carry = t / 10;
cur.next = new ListNode(t % 10);
cur = cur.next;
l1 = l1 == null ? null : l1.next;
l2 = l2 == null ? null : l2.next;
}
return dummy.next;
}
}
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