LeetCode 2: Add Two Numbers

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Add Two Numbers

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 contain a single digit. Add the two numbers and return it as a linked list.

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

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

第一次尝试: 50ms

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

        if(l1 == null)
            return l2;
        if(l2 == null)
            return l1;

        int sum = l1.val + l2.val;

        ListNode result = new ListNode(sum%10);

        if(sum > 9)
            result.next = addOne(addTwoNumbers(l1.next,l2.next));
        else
            result.next = addTwoNumbers(l1.next,l2.next); 

        return result;      
    }

    public ListNode addOne(ListNode l){
        if(l == null ){
            return new ListNode(1);
        }

        int sum = l.val + 1;

        ListNode result = new ListNode(sum%10);

        if(sum > 9)
            result.next = addOne(l.next);
        else
            result.next = l.next;

        return result;
    }
}

第二次尝试：（45ms）

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

        return addTwoNumbers(l1,l2,0);
    }

    public ListNode addTwoNumbers(ListNode l1, ListNode l2, int flag) {

        int val1,val2;

        if(l1 == null && l2 == null){
            if(flag == 1)
                return new ListNode(1);
            else
                return null;
        }
        if(l1 == null){
            if( flag == 0)
                return l2;
            else 
                return addTwoNumbers(l2, new ListNode(1),0);              
        } 

        if(l2 == null){
            if( flag == 0)
                return l1;
            else 
                return addTwoNumbers(l1, new ListNode(1),0);            
        } 

        int sum = l1.val + l2.val + flag;

        ListNode result = new ListNode(sum%10);

        if(sum > 9)
            result.next = addTwoNumbers(l1.next,l2.next,1);
        else 
            result.next = addTwoNumbers(l1.next,l2.next,0);

        return result;         
    } 
}

果然递归就是慢了点，看看大佬一遍扫描的写法：

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode prev = new ListNode(0);
        ListNode head = prev;
        int carry = 0;
        while (l1 != null || l2 != null || carry != 0) {
            ListNode cur = new ListNode(0);
            int sum = ((l2 == null) ? 0 : l2.val) + ((l1 == null) ? 0 : l1.val) + carry;
            cur.val = sum % 10;
            carry = sum / 10;
            prev.next = cur;
            prev = cur;

            l1 = (l1 == null) ? l1 : l1.next;
            l2 = (l2 == null) ? l2 : l2.next;
        }
        return head.next;
    }
}

题目地址：
https://leetcode.com/problems/add-two-numbers/description/