[LeetCode] 2. Add Two Numbers / Swift

[문제 보기]

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.

 

https://leetcode.com/problems/add-two-numbers/

Add Two Numbers - LeetCode

 

---

첫번째 방법

// 첫번째 버전

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        var cur1 = l1
        var cur2 = l2
        var buffer = 0
        let start = ListNode()
        var ansCur = start
        while cur1 != nil && cur2 != nil {
            ansCur.next = ListNode()
            ansCur = ansCur.next!
            let val = cur1!.val + cur2!.val + buffer
            ansCur.val = val % 10
            buffer = val / 10
            
            cur1 = cur1?.next
            cur2 = cur2?.next
        }
        
        if cur1 == nil && cur2 == nil && buffer == 0 {
            return start.next
        }
        
        var p = cur1
        if p == nil {
            p = cur2
        }
        
        while p != nil || buffer > 0 {
            let val = (p?.val ?? 0) + buffer
            ansCur.next = ListNode(val % 10)
            
            ansCur = ansCur.next!
            p = p?.next
            buffer = val / 10
        }
        
        return start.next
    }
}

 

첫번째 방법에서 수정한 방법:

기존 방법: while문에서 두개의 커서가 모두 nil이 아닐때까지 반복

수정한 방법: while문에서 두개의 커서가 모두 nil이 될때까지 반복. 이 방법은 while문 종료후에 nil이 아닌 어느 한쪽을 따로 이어주는 부분이 없음. 수행시간은 두번째 방법(52ms)보다 첫번째 방법(44ms)이 더 좋았음

// 두번째 버전

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        var cur1 = l1
        var cur2 = l2
        var buffer = 0
        let start = ListNode()
        var ansCur = start
        while cur1 != nil || cur2 != nil {
            ansCur.next = ListNode()
            ansCur = ansCur.next!
            let val = (cur1?.val ?? 0) + (cur2?.val ?? 0) + buffer
            ansCur.val = val % 10
            buffer = val / 10
            
            cur1 = cur1?.next
            cur2 = cur2?.next
        }
        
        if buffer > 0 {
            ansCur.next = ListNode(buffer)
        }
        
        return start.next
    }
}