LEETCODE #2 Add Two Numbers

最新推荐文章于 2021-11-03 13:44:22 发布

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本文详细解析了LeetCode上经典题目“两数相加”的解题思路，采用C++和Java语言实现，通过链表存储逆序数字并进行逐位相加，处理进位操作，适用于大整数加法练习。

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LEETCODE #2

问题描述

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.

0、背景

来源Leetcode题库Algorithms part，难度定级Medium。
参考Grandyang博客:Add Two Numbers 两数相加，语言选用C++，Java和Python。

1、算法

给定两个纯数字串，按位求和，注意进位，实际上就是大整数相加的模型，用于练习单链表的简单实现，C++的链表比Java难理解一点。注意最高位进位即可。

2、实现

a.CPP

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode *ans = new ListNode(-1);
        ListNode *temp = ans;
        int bitt = 0;
        while (l1 || l2)
        {
            int n1,n2;
            if(l1) n1 = l1->val; else n1 = 0;
            if(l2) n2 = l2->val; else n2 = 0;
            int sum = n1 + n2 + bitt;
            bitt = sum / 10;
            temp->next = new ListNode(sum%10);
            temp = temp->next;
            if (l1) l1 = l1->next;
            if (l2) l2 = l2->next;
        }
        if (bitt) temp->next = new ListNode(1);
        return ans->next;
        
    }
};

b.Java

/**
 * 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) {
        ListNode ans = new ListNode(-1);
        ListNode temp = ans;
        int bitt = 0;
        while (l1 != null || l2 != null)
        {
            int n1,n2;
            if (l1 != null) n1 = l1.val; else n1 = 0;
            if (l2 != null) n2 = l2.val; else n2 = 0;
            int sum = n1 + n2 + bitt;
            bitt = sum / 10;
            temp.next = new ListNode(sum%10);
            temp = temp.next;
            if (l1 != null) l1 = l1.next;
            if (l2 != null) l2 = l2.next;
        }
        if (bitt == 1) temp.next = new ListNode(1);
        return ans.next;
    }
}