第一个是测试用例代码，测试的是两个带头的逆序链表相加，并且有反转操作

但是题目要求的是不带头链表直接相加，不需要逆转，输出结果也是逆序的，

题解放在第二个代码中

#include<stdio.h>
#include<stdlib.h>
typedef struct ListNode {int val;struct ListNode* next;
}List;List* reverse(List* L) {List* p = L;List* q = NULL;while (p != NULL) {List* temp = p->next;p->next = q;q = p;p = temp;}List* head = (List*)(malloc)(sizeof(List));head->next = q;List* temp = head;while (temp->next->next != NULL) {temp = temp->next;}temp->next = NULL;return head;
}
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {List*L1=reverse(l1);L1->val = 0;List*L2=reverse(l2);L2->val = 0;List* L3 = (List*)(malloc)(sizeof(List));L3->next = NULL;L3->val = 0;int flag = 0;while (L1 != NULL || L2 != NULL) {List* temp = (List*)(malloc)(sizeof(List));temp->next = NULL;temp->val = 0;if (L1) {temp->val += L1->val;L1 = L1->next;}if (L2) {temp->val += L2->val;L2 = L2->next;}if (flag == 1) {temp->val += 1;flag = 0;}//进位if (temp->val >= 10) {temp->val= temp->val % 10;flag = 1;}//大于10进位List* cur = L3;while (cur->next != NULL) {cur = cur->next;}cur->next = temp;if (L1 == NULL && L2 == NULL && flag == 1) {List* digital = (List*)(malloc)(sizeof(List));digital->val = 1;digital->next = NULL;temp->next = digital;}}L3->next->val = 0;return reverse(L3->next);
}
List* makeList() {List* L = (List*)(malloc)(sizeof(List));L->next = NULL;int num;while (1) {scanf("%d", &num);if (num == 0) {break;}List* temp = (List*)(malloc)(sizeof(List));temp->next = NULL;temp->val = num;List* p = L;while (p->next != NULL) {p = p->next;}p->next = temp;}return L;
}
void print_List(List* L) {L = L->next; // 跳过头节点while (L != NULL) {printf("%d ", L->val);L = L->next;}printf("\n");
}int main() {List* L1 = makeList();printf("打印L1\n");print_List(L1);List* L3=reverse(L1);List* L2= makeList();printf("打印L2\n");print_List(L2);List* L4 = reverse(L2);printf("打印L3\n");print_List(L3);printf("打印L4\n");print_List(L4);List* L5 = addTwoNumbers(L3, L4);printf("打印L5\n");print_List(L5);
}

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {struct ListNode* L3 = (struct ListNode*)(malloc)(sizeof(struct ListNode));L3->next = NULL;L3->val=0;int flag = 0;while (l1 != NULL || l2 != NULL) {struct ListNode* temp = (struct ListNode*)(malloc)(sizeof(struct ListNode));temp->next = NULL;temp->val = 0;if (l1) {temp->val += l1->val;l1 = l1->next;}if (l2) {temp->val += l2->val;l2 = l2->next;}if (flag == 1) {temp->val += 1;flag = 0;}//进位if (temp->val >= 10) {temp->val= temp->val % 10;flag = 1;}//大于10进位struct ListNode* cur = L3;while (cur->next != NULL) {cur = cur->next;}cur->next = temp;if (l1 == NULL && l2 == NULL && flag == 1) {struct ListNode* digital = (struct ListNode*)(malloc)(sizeof(struct ListNode));digital->val = 1;digital->next = NULL;temp->next = digital;}}return (L3->next);
}