LeetCodeOJ_002: Add Two Numbers

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

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题目

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Total Accepted: 293186
Total Submissions: 1071927
Difficulty: Medium
Contributor: LeetCode

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

AC代码

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

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int carry = 0;
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        while (l1 != null || l2 != null) {
            int sum = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
            carry = sum / 10;
            cur.next = new ListNode(sum % 10);
            cur = cur.next;
            l1 = l1 == null ? null : l1.next;
            l2 = l2 == null ? null : l2.next;
        }
        if (carry != 0) cur.next = new ListNode(carry);
        return dummy.next;
    }
}

Complexity Analysis

• Time complexity : O(max(m, n)). Assume that mm and n represents the length of l1 and l2 respectively, the algorithm above iterates at most max(m, n)times.

• Space complexity : O(max(m, n)) The length of the new list is at most max(m,n) + 1 .

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这是我的AC代码

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

        ListNode head = new ListNode(0);
        ListNode point = head;
        int carry = 0;
        while(l1 != null && l2!=null){
            int sum = carry + l1.val + l2.val;
            point.next = new ListNode(sum % 10);
            carry = sum / 10;
            l1 = l1.next;
            l2 = l2.next;
            point = point.next;
        }

        while(l1 != null) {
            int sum =  carry + l1.val;
            point.next = new ListNode(sum % 10);
            carry = sum /10;
            l1 = l1.next;
            point = point.next;
        }

        while(l2 != null) {
            int sum =  carry + l2.val;
            point.next = new ListNode(sum % 10);
            carry = sum /10;
            l2 = l2.next;
            point = point.next;
        }

        if (carry != 0) {
            point.next = new ListNode(carry);
        }
        return head.next; 
    }
}