[刷题防痴呆] 0002 - 两数相加 (Add Two Num

题目地址

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

题目描述

2. 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.

思路

• 此题数字的链表表现形式是反着的. 所以逐位相加, 带上进位即可.
• dummy的应用.
• 注意进位的使用.
• while l1 != null && l2 != null
• while l1 != null
• while l2 != null
• if carry > 0
• 全部判断完毕. 返回dummy.next.

关键点

代码

• 语言支持：Java

/**
 * 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) {
        if (l1 == null || l2 == null) {
            return null;
        }
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        int carry = 0;
        
        while (l1 != null && l2 != null) {
            int sum = l1.val + l2.val + carry;
            cur.next = new ListNode(sum % 10);
            carry = sum / 10;
            cur = cur.next;
            l1 = l1.next;
            l2 = l2.next;
        }
        
        while (l1 != null) {
            int sum = l1.val + carry;
            cur.next = new ListNode(sum % 10);
            carry = sum / 10;
            cur = cur.next;
            l1 = l1.next;
        }
        
        while (l2 != null) {
            int sum = l2.val + carry;
            cur.next = new ListNode(sum % 10);
            carry = sum / 10;
            cur = cur.next;
            l2 = l2.next;
        }
        
        if (carry != 0) {
            cur.next = new ListNode(carry);
        }
        
        return dummy.next;
    }
}