445. Add Two Numbers II

Medium (Trung bình) Python 🔗 Xem trên LeetCode

📋 Đề Bài

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

 

Example 1:

Input: l1 = [7,2,4,3], l2 = [5,6,4]
Output: [7,8,0,7]

Example 2:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [8,0,7]

Example 3:

Input: l1 = [0], l2 = [0]
Output: [0]

 

Constraints:

  • The number of nodes in each linked list is in the range [1, 100].
  • 0 <= Node.val <= 9
  • It is guaranteed that the list represents a number that does not have leading zeros.

 

Follow up: Could you solve it without reversing the input lists?

🧠 Thuật Toán & Kỹ Thuật

Linked List (Danh sách liên kết)
⏱️ Thời gian O(n)
💾 Không gian O(n)

💻 Lời Giải

Python 0445-add-two-numbers-ii.py
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head:
            return None
        
        currNode = head
        nextNode = head.next
        prevNode = None
        
        while currNode and nextNode:
            currNode.next = prevNode
            prevNode = currNode
            currNode = nextNode
            nextNode = nextNode.next
            
        currNode.next = prevNode
        prevNode = currNode
            
        return prevNode
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        l1 = self.reverseList(l1)
        l2 = self.reverseList(l2)
        
        l3 = ListNode(0)
        cr = l3
        carry = 0
        
        while l1 or l2 or carry:
            total = carry
            if l1:
                total += l1.val
                l1 = l1.next
            if l2:
                total += l2.val
                l2 = l2.next
            cr.next = ListNode(total % 10)
            cr = cr.next
            carry = total // 10
            
        return self.reverseList(l3.next)