LeetCode 2. Add Two Numbers (Java)

tags: leetcode Java

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.

有兩個非空的linked list代表的整數，其中每個節點包含一個數字。數字存儲按照在原來整數中相反的順序。寫出一個函數將兩個整數相加，用linked list形式回傳兩者的合。

Example

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.

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

Idea

• 會有進位問題所以設定一個carry變數（0->沒有進位，1->有進位）
• 並且在計算完一個節點之後curr中指向result下一個節點在做同樣的計算，l1 l2則判斷是否下一個節點是否NULL也指向下一個節點
• 直到l1和l2都指向連結的最後NULL則程式結束。

Code

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode result = new ListNode(0);//設定新的空鏈結儲存結果
        ListNode curr = result;
        int carry = 0;//進位
        
        while(l1 != null || l2 != null) {
            int x = (l1!=null) ? l1.val: 0;
            int y = (l2!=null) ? l2.val: 0;
            int sum = x + y + carry;
            
            if(sum >= 10) carry = 1;
            else carry = 0;
            
            curr.next = new ListNode(sum%10);
            curr = curr.next;
            if(l1 != null) l1 = l1.next;
            if(l2 != null) l2 = l2.next;
        }
        //carry = 1
        if(carry == 1) curr.next = new ListNode(1);
        
        return result.next;//回傳鏈結的第一個節點
    }
}