题解 | #链表相加(二)#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param head1 ListNode类
 * @param head2 ListNode类
 * @return ListNode类
 */

#include <stdlib.h>
struct ListNode* ReverseList(struct ListNode* head ) {
    // write code here
    struct ListNode* next = NULL;
    struct ListNode* prev = NULL;
    struct ListNode* curr = head;

    while (curr != NULL) {
        next = curr -> next;
        curr -> next = prev;
        prev = curr;
        curr = next;
    }

    return prev;
}

struct ListNode* addInList(struct ListNode* head1, struct ListNode* head2 ) {
    // write code here
    if (head1 == NULL)
        return head2;
    if (head2 == NULL)
        return head1;

    head1 = ReverseList(head1);
    head2 = ReverseList(head2);
    int temp = 0;
    struct ListNode* numList = malloc(sizeof(struct ListNode*));
    struct ListNode* head = numList;
    int val1, val2, num;

    while (head1 != NULL || head2 != NULL || temp != 0) {
        val1 = head1 == NULL ? 0 : head1 -> val;
        val2 = head2 == NULL ? 0 : head2 -> val;

        num = val1 + val2 + temp;
        temp = num / 10;
        num %= 10;

        head -> next = malloc(sizeof(struct ListNode*));
        head -> next -> val = num;
        head = head -> next;

        if (head1 != NULL)
            head1 = head1->next;
        if (head2 != NULL)
            head2 = head2->next;
    }

    return  ReverseList(numList -> next);

}