Reverse a singly linked list.

解题思路

对于非递归实现，思路是依次将从第二个结点到最后一个结点的后继设为头结点，然后将该节点设为头结点（需记住将原头结点的后继设为空）。
对于递归实现，首先反转从第二个结点到最后一个结点的链表，然后再将头结点放到已反转链表的最后，函数返回新链表的头结点。

非递归实现代码1

//Runtime:10 ms
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (head == NULL) return NULL;
        ListNode *pre = head;
        ListNode *cur = head->next;
        while (cur != NULL)
        {
            pre->next = cur->next;
            cur->next = head;
            head = cur;
            cur = pre->next;
        }

        return head;
    }
};

非递归实现代码2

//Runtime:22 ms
class Solution{
public:
    ListNode* reverseList(ListNode* head){
        if (head == NULL) return NULL;
        ListNode *cur = head->next;
        head->next = NULL;
        while (cur != NULL)
        {
            ListNode *temp = cur->next;
            cur->next = head;
            head = cur;
            cur = temp;
        }

        return head;
    }
};

递归实现代码

//Runtime:10 ms
class Solution{
public:
    ListNode* reverseList(ListNode* head){
        //此处的条件不能写成if(head == NULL)
        if (head == NULL || head->next == NULL) return head;
        ListNode *newhead = reverseList(head->next);
        head->next->next = head;
        head->next = NULL;

        return newhead;
    }
};

测试代码如下：

void printList(ListNode *head)
{
    while(head != NULL)
    {
        cout<<head->val<<" ";
        head = head->next;
    }
    cout<<endl;
}

void dropList(ListNode *head)
{
    if (head == NULL) return;
    ListNode *temp;
    while (head != NULL)
    {
        temp = head->next;
        delete head;
        head = temp;
    }
}

int main()
{
    ListNode *head = new ListNode(0);
    ListNode *cur = head;
    for (int i = 0; i < 10; i++)
    {
        ListNode *newnode = new ListNode(floor(rand()%100));
        cur->next = newnode;
        cur = newnode;
    }
    printList(head);
    Solution s;
    head = s.reverseList(head);
    printList(head);
    dropList(head);
}