classSolution {publicListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode l3=null;booleanadd =false;while(l1 !=null|| l2 !=null) {//位数相加intplus;if(l1 ==null) { plus=l2.val; }elseif(l2 ==null) { plus=l1.val; }else{ plus= l1.val +l2.val; }if(add) { plus++; ...
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, exc...
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, exc...
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. 你有两个非空链表,表示两个非负整数,数字以相反的顺序存储,每个节点中包...
You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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, excep...
* 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,...
* type ListNode struct { * Val int * Next *ListNode * } */ func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode { // 哨兵结点,方便后续处理 head_pre := &ListNode{} // 结果链表的尾结点,方便用尾插法插入 tail := head_pre // 进位值,初始化为 0 carry := 0 // 如果两个链表...
classSolution(object):defaddTwoNumbers(self,l1,l2):""":type l1: ListNode:type l2: ListNode:rtype: ListNode"""carry=0cur=dummy=ListNode(0)#遍历l1, l2 链表,并依次赋值给cur 节点whilel1orl2:ifl1andl2:ifl1.val+l2.val+carry>=10:cur.next=ListNode((l1.val+l2.val+carry)%10)carry=1els...
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...
检查incry是否不为0,如果是,则创建值为incry的节点,result.next指向这个节点。 返回head.next节点。 Talk is cheap, show me the code! class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode p1 = l1; ListNode p2 = l2; ...