【LeetCode】33.Linked List — Add Two Numbers两数相加 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 assum...
【LeetCode OJ】Add Two Numbers 题目:You are given two linked lists representing two non-negative numbers. 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. Input: (2 -> 4 -> 3) + (5 -> 6 ...
题目 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...
需要注意的地方是考虑进位,中间进位和最后一位进位,我的解决方法是先不考虑进位,最后再一起考虑进位。 # Definition for singly-linked list.# class ListNode:# def __init__(self, val=0, next=None):# self.val = val# self.next = nextclassSolution:defaddTwoNumbers(self,l1:ListNode,l2:ListNode)->...
# 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: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: # 哨兵结点,方便后续处理 head_pre =...
Leetcode 2. 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 contain a single digit. Add the two numbers and return it as a linked list....
LeetCode: 2. Add Two Numbers LeetCode: 2. Add Two Numbers 题目描述 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...
3 输入与输出：/*** 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) { }};4 解决思路：从表头开始相加，记录每次相加...
class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { int sum = l1->val + l2->val; int value = sum % 10, c = sum / 10; ListNode *res = new ListNode(value); ListNode *p1 = l1, *p2 = l2, *pr = res; ...
方法二:遍历两个list,直接按位计算,记录是否进位# 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...