【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...
/*** Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * }*/classSolution {publicListNode addTwoNumbers(ListNode l1, ListNode l2) { Stack<Integer> s1 =newStack<Integer>(); Stack<Integer> s2 =newStack<Integer>...
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题解---2】Add Two Numbers 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....
LeetCode-Java-2. Add Two Numbers 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....
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; ...
* Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */classSolution{public:ListNode*addTwoNumbers(ListNode*l1,ListNode*l2){ListNode*pRoot=NULL;do{if(l1==NULL){pRoot=l2;break;}if(l2==NULL){pRoo...