每日一题《leetcode– LCR 025.两数相加||》

作者 : admin 本文共1147个字,预计阅读时间需要3分钟 发布时间: 2024-06-2 共3人阅读

http://leetcode.cn/problems/lMSNwu/

每日一题《leetcode– LCR 025.两数相加||》插图

每日一题《leetcode– LCR 025.两数相加||》插图(1)


分别把给定的两个链表翻转,然后从头开始相加。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
//反转链表
struct ListNode* reverselist(struct ListNode*h1)
{
    struct ListNode* newhead = NULL;
    struct ListNode* cur = h1;

    while(cur)
    {
        struct ListNode* next = cur->next;

        cur->next = newhead;
        newhead = cur;
        cur = next;
    }

    return newhead;
}

struct ListNode* add(struct ListNode*head1 , struct ListNode* head2)
{   
    //返回的新链表的头和尾
    struct ListNode* head = NULL,* tail = NULL;
    //进位
    int carry = 0;

    while(head1 || head2)
    {
        int n1 = head1? head1->val : 0;
        int n2 = head2? head2->val : 0;


        int sum = n1 + n2 + carry;

        if(head == NULL)
        {
            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(head1)
        {
            head1 = head1->next;
        }

        if(head2)
        {
            head2 = head2->next;
        }
    }

    if(carry > 0)
    {
        tail->next = (struct ListNode*)malloc(sizeof(struct ListNode));
        tail->next->val = carry;
        tail = tail->next;
        tail->next = NULL;
    }


    return head;
}

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
    l1 = reverselist(l1);
    l2 = reverselist(l2);

    struct ListNode* l3 = add(l1,l2);
    return reverselist(l3);
}

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