【Leetcode -1290.二进制链表转整数 -剑指Offer 06.从尾到头打印链表】

		int getDecimalValue(struct ListNode* head)
		{
		    //定义sum为0，由于链表中的值不是0就是1
		    //所以每次都用sum的最低位0去按位或上head的val，即可得到val
		    //每次按位或完，head往后走；sum向左移，右边会补0
		    int sum = 0;
		    while (head)
		    {
		        sum <<= 1;
		        sum |= head->val;
		        head = head->next;
		    }
		    return sum;
		}

		int* reversePrint(struct ListNode* head, int* returnSize)
		{
		    //链表为空，长度返回0，返回空
		    if (head == NULL)
		    {
		        *returnSize = 0;
		        return NULL;
		    }
		
		    //count计算链表长度
		    //反转链表
		    int count = 0;
		    struct ListNode* curr = head, * prev = NULL;
		    while (curr)
		    {
		        struct ListNode* next = curr->next;
		        curr->next = prev;
		        prev = curr;
		        curr = next;
		        count++;
		    }
		
		    //开辟一个数组返回
		    //更新返回长度
		    int* p = (int*)malloc(sizeof(int) * count);
		    *returnSize = count;
		
		    //将反转后链表中的val放到数组中
		    for (int i = 0; i < count; i++)
		    {
		        p[i] = prev->val;
		        prev = prev->next;
		    }
		
		    return p;
		}