每日一题——两个链表相加

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 * };
 */
/**
 * 
 * @param head1 ListNode类 
 * @param head2 ListNode类 
 * @return ListNode类
 */
struct ListNode* FilpLinkList(struct ListNode* phead)      //反转链表
{
    if(!phead)
        return NULL;
  struct ListNode* newHead = (struct ListNode*)malloc(sizeof(struct ListNode));
    struct ListNode* cur = phead;
    newHead->next = phead;
    while(cur->next)
    {
        struct ListNode* temp = cur->next;
        cur->next = temp->next;
        temp->next = newHead->next;
        newHead->next = temp;
    }
    return newHead->next;
}
struct ListNode* CreatNewNode(int num)    //创建新节点
{
    struct ListNode* newNode = (struct ListNode *)malloc(sizeof(struct ListNode));
    newNode->next = NULL;
    newNode->val = num;
    return newNode;
}
struct ListNode* addInList(struct ListNode* head1, struct ListNode* head2 ) {
    if(head1 == NULL || head2 == NULL)    //只要head1、head2有一个为空，就相当于加数为0，直接返回不为空的表头，两个都为空，同理。
        return head1 == NULL ? head2 : head1;
    struct ListNode* head3 = (struct ListNode*)malloc(sizeof(struct ListNode)); //新建表头
    //反转链表
    head1 = FilpLinkList(head1);    
    head2 = FilpLinkList(head2);
    struct ListNode* cur1 = head1, *cur2 = head2,*cur3 = head3;
    int flag = 0,val1,val2;
    while(cur1 || cur2 || flag)
    {
        //如果cur为空，说明已经走到表尾，直接令val为0即可
        val1 = cur1 == NULL ? 0 : cur1->val;
        val2 = cur2 == NULL ? 0 : cur2->val;
        cur3->next = CreatNewNode((val1 + val2 + flag) % 10);   //创建新节点
        flag = (val1 + val2 + flag) / 10;   //更新flag
        cur3 =cur3->next;
        //如果cur为空，则停止下滑
        cur1 = cur1 == NULL ? NULL : cur1->next;
        cur2 = cur2 == NULL ? NULL : cur2->next;
    }
    return FilpLinkList(head3->next); //翻转新表
}