LeetCode 0002. Add Two Numbers Solution Java | Python | C++ | JavaScript + Others

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0002. Add Two Numbers

Description

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 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 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

Example 2:

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

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]

 

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.

Solutions

Solution 1: Simulation

We traverse two linked lists $l_1$ and $l_2$ at the same time, and use the variable $carry$ to indicate whether there is a carry.

Each time we traverse, we take out the current bit of the corresponding linked list, calculate the sum with the carry $carry$, and then update the value of the carry. Then we add the current bit to the answer linked list. If both linked lists are traversed, and the carry is $0$, the traversal ends.

Finally, we return the head node of the answer linked list.

The time complexity is $O(\max (m, n))$, where $m$ and $n$ are the lengths of the two linked lists. We need to traverse the entire position of the two linked lists, and each position only needs $O(1)$ time. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

PythonJavaC++GoTypeScriptRustJavaScriptC#PHPSwiftRubyNimC
# 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]: dummy = ListNode() carry, curr = 0, dummy while l1 or l2 or carry: s = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carry carry, val = divmod(s, 10) curr.next = ListNode(val) curr = curr.next l1 = l1.next if l1 else None l2 = l2.next if l2 else None return dummy.next(code-box)

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