【LeetCode】2. 两数相加

题目链接

Python3

方法： 模拟 $⟮O(n)、O(1)⟯$

# 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]:cur = dummy = ListNode(None)carry = 0while l1 or l2:carry += (l1.val if l1 else 0) + (l2.val if l2 else 0 )cur.next = ListNode(carry % 10)  # 注意这里carry = carry // 10cur = cur.next if l1: l1 = l1.next if l2: l2 = l2.nextif carry > 0:  # 处理 最后一个进位cur.next = ListNode(carry) return dummy.next 

C++

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {ListNode *dummy = new ListNode(0);  // 虚拟结点 定义ListNode* cur = dummy;int carry = 0;while (l1 || l2){int n1 = l1 ? l1->val : 0;int n2 = l2 ? l2->val : 0;int total = n1 + n2 + carry;cur->next = new ListNode(total % 10);carry = total / 10;cur = cur->next;if (l1){l1 = l1->next;}if (l2){l2 = l2->next;}}if (carry > 0){cur->next = new ListNode(carry);}return dummy->next;}
};