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...
};structListNode* addTwoNumbers(structListNode* l1,structListNode*l2) {intdebug =0;structListNode* cur1 =l1;structListNode* cur2 =l2;structListNode* l3 =malloc(sizeof(structListNode)); l3->val =0; l3->next =NULL;structListNode* tmp =l3;structListNode* cur3 =l3;intv1 =0, v2 =0, ...
leetcode刷题: 002 Add Two Numbers 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....
4 解决思路：从表头开始相加，记录每次相加的进位。5 具体算法：伪代码如下：初始化进位变量carry=0将p和q分别初始化为l1和l2的头部。循环遍历列表l1和l2，直到达到两端。 将x设置为节点p的值。如果p已达到l1的末尾，则设置为0。 将y设置为节点q的值。如果q已达到l2的末尾，则设置为0。 设置sum =...
LeetCode 2. Add Two Numbers non-empty You may assume the two numbers do not contain any leading zero, except the number 0 itself. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: answer: class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {...
类似题目:LeetCode 67 - Add Binary | 二进制求和 (Rust) 时间复杂度:O(|l1| + |l2|) 需要遍历 l1 中的全部 O(|l1|) 个结点 需要遍历 l2 中的全部 O(|l2|) 个结点 空间复杂度:O(1) 需要为结果链表中的全部 O(max(|l1|, |l2|)) 个结点分配空间 (理论上可以复用已有的结点,这样就只需要定...
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...
{public:ListNode*addTwoNumbers(ListNode*l1,ListNode*l2){if(l1==nullptr){returnl2;}if(l2==nullptr){returnl1;}intsum=0;ListNode*head=newListNode(0);ListNode*cur=head;while(1){if(l1!=nullptr){sum+=l1->val;l1=l1->next;}if(l2!=nullptr){sum+=l2->val;l2=l2->next;}cur->val=sum%...
# 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 """ num1 = 0 scale = 1 while l1: num1 = num1 + l1.val * sc...
addTwoNumbers(l1, l2); 72 while (l != NULL){ 73 cout << l->val << endl; 74 l = l->next; 75 } 76 while (1); 77 } 运行结果: 最后实现不等长有进位的求和,即实现题目要求(注意最后一位有进位的情况) 代码语言:javascript 复制 1 #include <iostream> 2 3 using namespace std; 4 ...