题解 | #反转链表#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param head ListNode类
 * @return ListNode类
 */
struct ListNode* ReverseList(struct ListNode* head ) {
    // write code here
    struct ListNode* headtemp = NULL; 
    struct ListNode* headlist = NULL;
    int i = 0;

    while (head != NULL) {
        if (0 == i) {
            headtemp = (struct ListNode*)malloc(sizeof(struct ListNode));
            memcpy(headtemp, head, sizeof(struct ListNode));
            headlist = headtemp;
            headtemp->next = NULL;
            i++;
            printf("fxs i = %d,headtemp->val = %d \r\n", i, headtemp->val);

        } else {
            headtemp = (struct ListNode*)malloc(sizeof(struct ListNode));
            memcpy(headtemp, head, sizeof(struct ListNode));
            i++;
            headtemp->next = headlist;
            headlist = headtemp;
            printf("fxs i = %d,headtemp->val = %d \r\n", i, headtemp->val);
        }
        head = head->next;
    }

    headtemp = headlist;
    i = 0;
    while(headtemp != NULL)
    {
        i++;
        printf("i = %d,headtemp->val = %d \r\n", i, headtemp->val);
        headtemp = headtemp->next;
    }

    return headlist;
}