Input:
1->2->3

Output:
1->2->4

1，反转后从左向右加一，当最高位出现进位的时候，额外申请一个节点保存进位，之后在反转链表并返回。
2，当没有进位的适合不需要继续遍历链表
3，在遍历链表做加一的同时，记录下当前节点的前一个节点，当循环退出条件是p==NUll适合，同时又存在最高位有进位的情况，需要prev指针指向这个进位节点

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* reverseList(struct ListNode* head) {
    if(head == NULL)
        return NULL;
    if(head->next == NULL)
        return head;
    struct ListNode *node;

    node = reverseList(head->next);
    head->next->next = head;
    head->next = NULL;

    return node;    
}

struct ListNode* plusOne(struct ListNode* head) {
    
    struct ListNode *head1, *head2;
    struct ListNode *p,*prev;
    int val;
    if(head == NULL )
        return NULL;
    head1 = reverseList(head);
    p = head1;
    val = p->val +1;
    p->val = val%10;
    int carry = val/10;
    prev = p;
    p = p->next;
    while(p&&carry){
        val = p->val+carry;
        p->val = val%10;
        carry = val/10;
        prev = prev->next;
        p = p->next;
    }
   
    if(carry){
        printf("++");
        struct ListNode *node = malloc(sizeof(struct ListNode));
        node->val = carry;
        node->next = NULL;
        prev->next = node;
    }

    head2 = reverseList(head1);

    return head2;
}

struct ListNode* plusOne(struct ListNode* head) {
    
    struct ListNode *head1, *head2;
    struct ListNode *p,*prev;
    int val;
    if(head == NULL )
        return NULL;
    head1 = reverseList(head);
    p = head1;
    val = p->val +1;
    p->val = val%10;
    int carry = val/10;
    prev = p;
    p = p->next;
    while(p&&carry){
        val = p->val+carry;
        p->val = val%10;
        carry = val/10;
        prev = prev->next;
        p = p->next;
    }
   
    if(carry){
        printf("++");
        struct ListNode *node = malloc(sizeof(struct ListNode));
        node->val = carry;
        node->next = NULL;
        prev->next = node;
    }

    head2 = reverseList(head1);

    return head2;
}