Add the two numbers in the linked list of the second question of C language. Ergodic method

Here are two for you Non empty The linked list of , Represents two nonnegative integers . Each of them is based on The reverse Stored in , And each node can only store a Numbers .

Please add up the two numbers , And returns a linked list representing sum in the same form .

You can assume that in addition to the numbers 0 outside , Neither of these numbers 0 start .

Example 1：

Input ：l1 = [2, 4, 3], l2 = [5, 6, 4]
Output ：[7, 0, 8]
explain ：342 + 465 = 807.

Example 2：

Input ：l1 = [0], l2 = [0]
Output ：[0]

Example 3：

Input ：l1 = [9, 9, 9, 9, 9, 9, 9], l2 = [9, 9, 9, 9]
Output ：[8, 9, 9, 9, 0, 0, 0, 1]

source ： Power button （LeetCode）
link ：https ://leetcode.cn/problems/add-two-numbers
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Think of two linked lists as traversal of the same length , If a link list is short, fill it in the front 000, such as 987 + 23 = 987 + 023 = 1010
Each bit needs to consider the carry problem of the previous bit at the same time , The carry value needs to be updated after the current bit calculation
If the two lists are all traversed , Carry value is 111, Then add the node at the front of the new list 111
Tips ： For the list problem , When the return result is the header node , Usually you need to initialize a pre pointer first pre, The next node of the pointer points to the real header node head. The purpose of using the pre pointer is that there is no node value available when the list is initialized , And the construction process of linked list needs pointer movement , This will result in the loss of the head pointer , Unable to return a result

#define _CRT_SECURE_NO_WARNINGS

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

struct ListNode 
{
    int val;
    struct ListNode *next;
};
 
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2)
{
	int pos = 0;// Check whether to carry , Carry is 1 Non carry is 0
	struct ListNode* head = NULL;// Chain header pointer 
	struct ListNode* tail = NULL;// Pointer at the end of the list 
	struct ListNode* mid = NULL;//Bianli zhizhen
	int s1 = 0, s2 = 0;//s1 representative l1 Corresponding value ,s2 Corresponding l2 Corresponding value 
	int sum = 0	;
	while (l1 != NULL || l2 != NULL)// Cycle when both linked lists are not at the end 
		//[5]
		//[5]
	{
		if (l1 == NULL)
			s1 = 0;
		else if (l1 != NULL)
			s1 = l1->val;
		if (l2 == NULL)
			s2 = 0;
		else if (l2 != NULL)
			s2 = l2->val;
		sum = s1 + s2 + pos;
		if (sum>= 10)
			pos = 1;
		else if (sum < 10)
			pos = 0;
		if (head == NULL)
		{
			head = tail = (struct ListNode*)malloc(sizeof(struct ListNode));
			head->val = sum % 10;
			tail->next = NULL;
		}
		else if (head != NULL)
		{
			mid = (struct ListNode*)malloc(sizeof(struct ListNode));
			mid->val = sum%10;
			mid->next = tail->next;// Use tail insertion 
			tail->next = mid;
			tail = mid;
		}
		if (l1 != NULL)
			l1 = l1->next;
		if (l2 != NULL)
			l2 = l2->next;
	}
	if (pos == 1)
	{
		tail = (struct ListNode*)malloc(sizeof(struct ListNode));
		tail->val = pos;
		tail->next = NULL;
		if (head->next == NULL)// Judge mid Whether there is   If mid There is no direct to mid Operation may illegally access 
			head->next = tail;
		else
			mid->next = tail;
	}
	return head;
}