1.题目要求:

给你一个 非空 链表的头节点 head ，表示一个不含前导零的非负数整数。

将链表 翻倍 后，返回头节点 head 。

2.题目代码:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* doubleIt(struct ListNode* head){
    struct ListNode* cur = head;
    int count = 0;
    //1.遍历链表，得到结点个数
    while(cur){
        count++;
        cur = cur->next;
    }
    int* number = (int*)malloc(sizeof(int) * count);//2.根据结点个数用malloc申请数组
    int j = 0;
    cur = head;
    //3.把链表的结点放入数组
    while(cur){
        number[j] = cur->val;
        j++;
        cur = cur->next;
    }
    //4.给每个数组都乘以2
    for(int i = 0;i < j;i++){
        number[i] *= 2;
    }
    //5.进行进位
    for(int i = j - 1;i > 0;i--){
        if(number[i] > 9){
            number[i] %= 10;
            number[i - 1] += 1;
        }
    }
    //6.判断数组第一个数是否大于9
    if(number[0] > 9){
        struct ListNode* newnode = (struct ListNode*)malloc(sizeof(struct ListNode));
        newnode->val = number[0] / 10;
        newnode->next = head;
        head = newnode;
        cur = head;
        cur = cur->next;
        cur->val = number[0] % 10;
        int i = 1;
        cur = cur->next;
        while(cur){
            cur->val = number[i];
            i++;
            cur = cur->next;
        }
        return head;
    }else{
            cur = head;
            int i = 0;
            while(cur){
                cur->val = number[i];
                i++;
                cur = cur->next;
            }
            return head;
    }
}