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
 */
struct ListNode* reverseList(struct ListNode* head) {
    struct ListNode *prev = NULL;
    struct ListNode *current = head;
    while (current != NULL) {
        struct ListNode *nextNode = current->next;
        current->next = prev;
        prev = current;
        current = nextNode;
    }
    return prev;
}

struct ListNode* addInList(struct ListNode* head1, struct ListNode* head2 ) {
    // write code here
    if(head1 == NULL)
        return head2;
    else if(head2 == NULL)
        return head1;
    head1=reverseList(head1);
    head2=reverseList(head2);//反转链表
   
    struct ListNode *dummy = (struct ListNode*)malloc(sizeof(struct ListNode));
    dummy->val = 0;
    dummy->next = NULL;
    struct ListNode *current = dummy;
    int carry=0;

    while(head1 != NULL || head2 != NULL || carry !=0)
    {
        int sum=carry;

        if(head1 != NULL)
        {
            sum +=head1->val;
            head1=head1->next;
        }

        if(head2 != NULL)
        {
            sum +=head2->val;
            head2=head2->next;
        }

        carry = sum/10;
        sum=sum%10;

        current->next = (struct ListNode*)malloc(sizeof(struct ListNode));
        current->next->val = sum;
        current->next->next = NULL;

        current=current->next;
    }
    return  reverseList(dummy->next);

    
}