LeetCode 2.两数相加 每日一题

1、问题描述

给你两个 非空 的链表,表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的,并且每个节点只能存储 一位 数字。

请你将两个数相加,并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外,这两个数都不会以 0 开头。

示例 1:
输入:l1 = [2,4,3], l2 = [5,6,4]
输出:[7,0,8]
解释:342 + 465 = 807.
示例 2:

输入:l1 = [0], l2 = [0]
输出:[0]
示例 3:

输入:l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出:[8,9,9,9,0,0,0,1]

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/add-two-numbers

2、C语言实现

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode* head = NULL;
    struct ListNode* tail = NULL;
    int carry = 0;
    while(l1 || l2)
    {
        int n1 = l1 ? l1->val : 0;
        int n2 = l2 ? l2->val : 0;
        int sum = n1 + n2 + carry;
        if(!head)
        {
            head = tail = (struct ListNode*)malloc(sizeof(struct ListNode));
            tail->val = sum % 10;
            tail->next = NULL;
        }
        else 
        {
            tail->next = (struct ListNode*)malloc(sizeof(struct ListNode));
            tail->next->val = sum % 10;
            tail = tail->next;
            tail->next = NULL;
        }
        carry = sum / 10;
        if(l1)
        {
            l1 = l1->next;
        }
        if(l2)
        {
            l2 = l2->next;
        }
    }
    if(carry > 0)
    {
        tail->next = (struct ListNode*)malloc(sizeof(struct ListNode));
        tail->next->val = carry;
        tail->next->next = NULL;
    }
    return head;
}

3、Java实现

/**
 * 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 head = null;
        ListNode tail = null;
        int carry = 0;
        while(l1 != null || l2 != null){
            int n1 = l1 != null ? l1.val : 0;
            int n2 = l2 != null ? l2.val : 0;
            int sum = n1 + n2 + carry;
            if(head == null){
                head = tail = new ListNode(sum % 10);
            }else {
                tail.next = new ListNode(sum % 10);
                tail = tail.next;
            }
            carry = sum / 10;
            
            if(l1 != null){
                l1 = l1.next;
            }
            if(l2 != null){
                l2 = l2.next;
            }
        }
        if(carry > 0){
            tail.next = new ListNode(carry);
        }
        return head;
    }
}

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