【Leetcode】2. 两数相加

题目

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。

提示

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]

代码

#include <string.h>
#include <stdlib.h>
#include <stdio.h>

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode {
    int val;
    struct ListNode *next;
};

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    // Notice! when init the point the * is before the value
    // struct ListNode *head = NULL, *tail = NULL;
    struct ListNode* head = NULL;
    struct ListNode* tail = NULL;
    int carry = 0;
    while(l1 || l2) {
        int num1 = l1 ? l1->val : 0;
        int num2 = l2 ? l2->val : 0;
        int sum  = num1 + num2 + carry;
        if(!head) {
            // init the head point to keep the point of first node
            head = tail = (struct ListNode*)malloc(sizeof(struct ListNode));
            tail->val  = sum % 10;
            tail->next = NULL;
        } else {
            // init the next node
            tail->next       = (struct ListNode*)malloc(sizeof(struct ListNode));
            tail->next->val  = sum % 10;
            tail->next->next = NULL;
            // move the point
            tail = tail->next;
        }
        carry = sum / 10;
        if (l1) {
            l1 = l1->next;
        }
        if (l2) {
            l2 = l2->next;
        }
    }
    // if the carry not equal to zero, we need another node to store it.
    if (carry > 0) {
        tail->next       = (struct ListNode*)malloc(sizeof(struct ListNode));
        tail->next->val  = carry;
        tail->next->next = NULL;
    }
    return head;
}

struct ListNode* createLinkedList(int arr[], int n) {
    struct ListNode* head = NULL;
    struct ListNode* tail = NULL;
    for (int i = 0; i < n; i++) {
        if (!head) {
            head = tail = (struct ListNode*)malloc(sizeof(struct ListNode));
            tail->val = arr[i];
            tail->next = NULL;
        } else {
            tail->next = (struct ListNode*)malloc(sizeof(struct ListNode));
            tail->next->val = arr[i];
            tail->next->next = NULL;
            tail = tail->next;
        }
    }
    return head;
}

int main() {
    struct ListNode* l1 = (struct ListNode*)malloc(sizeof(struct ListNode));
    struct ListNode* l2 = (struct ListNode*)malloc(sizeof(struct ListNode));
    // init the list
    int arr[] = {2,4,3};
    int arr2[] = {5,6,4};
    l1 = createLinkedList(arr, 3);
    l2 = createLinkedList(arr2, 3);
    struct ListNode* res = addTwoNumbers(l1, l2);
    // print the list
    while(res) {
        printf("%d\n", res->val);
        res = res->next;
    }
    return 0;
}

链表问题大多考察链表操作，所以没有分析
这个题注意，给出的链表和最后的结果都是逆序，直接向后进位即可