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

    if ((NULL == head) || (NULL == head->next)) {
        return head;
    }

    struct ListNode* temphead = (struct ListNode*)malloc(sizeof(struct ListNode));
    temphead->next = head;
    struct ListNode* temp1 = head;
    struct ListNode* temp2 = head->next;

    while (temp2 != NULL) {
        temp1->next = temp2->next;//连
        //temp2->next = temp1;
        temp2->next = temphead->next;//掉
        temphead->next = temp2;//接
        printf("temp2 = %d\n", temp2->val);

        temp2 = temp1->next;//移
    }

    head = temphead->next;
    free(temphead);
    return head;


}

此链表无论头节点是否有数据，都可以改变此用例来实现链表反转