剑指 Offer 06. 从尾到头打印链表

输入一个链表的头节点，从尾到头反过来返回每个节点的值（用数组返回）。

示例 1：

输入：head = [1,3,2]
输出：[2,3,1]

限制：

0 <= 链表长度 <= 10000

作者：Krahets
链接：https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5dt66m/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
通过先把链表逆序，在逆序过程中同时统计链表节点的数量；申请链表节点数量的数据，然后再依次遍历链表，将链表中的节点值域的值赋值给数据；

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode* listReverse(struct ListNode *head, int *returnSize)
{
    if (!head)
    {
        *returnSize = 0;
        return NULL;
    }

    struct ListNode *pre = head;
    struct ListNode *cur = pre->next;
    struct ListNode *nex;
    int NodeNum = 1;
    while(cur)
    {
        nex = cur->next;
        cur->next = pre;
        pre = cur;
        cur = nex;
        NodeNum ++;
    }

    head->next = NULL;

    head = pre;

    *returnSize = NodeNum;

    return head;
}

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* reversePrint(struct ListNode* head, int* returnSize){
    struct ListNode *pf = listReverse(head, returnSize);
    int i = 0;

    printf("*returnSize:%d\n", *returnSize);

    int *res = (int *)malloc(sizeof(int) * (*returnSize));
    while(pf) {
        res[i++] = pf->val;
        pf = pf->next;
    }

    return res;
}