leecode206_反转链表

第一种思路，比较冗杂 

struct ListNode* reverseList(struct ListNode* head)
{
    struct ListNode* guard = (struct ListNode*)malloc(sizeof(struct ListNode));
    guard->next = NULL;
    struct ListNode* cur, * tail, * tmp;
    cur = tail = head;
    while (cur)
    {
        if (tail == head)
        {
            guard->next = tail;
            cur = cur->next;
            tail->next = NULL;
        }
        else
        {
            guard->next = tail;
            cur = cur->next;
            tail->next = tmp;
        }
        tmp = tail;
        tail = cur;
    }
    return guard->next;
    free(guard);
}

第二种思路

struct ListNode* reverseList(struct ListNode* head)
{
    struct ListNode* rhead, * cur, * next;
    rhead = NULL;
    cur = next = head;
    while (cur != NULL)
    {
        next = cur->next;
        cur->next = rhead;

        rhead = cur;
        cur = next;
    }
    return rhead;
}

第三种，较优

struct ListNode* reverseList(struct ListNode* head)
{
    if (head == NULL)
    {
        return NULL;
    }
    struct ListNode* n1, * n2, * n3;
    n1 = NULL;
    n2 = head;
    n3 = n2->next;

    while (n2)
    {
        n2->next = n1;
        n1 = n2;
        n2 = n3;
        if (n3 != NULL)
            n3 = n3->next;

    }
    return n1;

}