LeetCode OJ：Reverse Linked List （反转链表）

Reverse a singly linked list.

做II之前应该先来做1的，这个倒是很简单，基本上不用考虑什么，简单的链表反转而已：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(head == NULL) return NULL;
        ListNode * prev = NULL;
        ListNode * p = head;
        ListNode * next;
        while(p != NULL){
            next = p->next;
            p->next = prev;
            prev = p;
            p = next;
        }
        return prev;
    }
};

 java版本的如下所示，基本上没什么区别，就是普通解法：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseList(ListNode head) {
        if(head == null)return null;
        ListNode helper = new ListNode(-1);
        helper.next = head;
        ListNode p = helper;
        ListNode pPre = helper.next;
        ListNode tmp = null;
        while(p.next!=null){
            tmp = p.next;
            p.next = pPre;
            pPre = p;
            p = tmp;
        }
        p.next = pPre;
        helper.next.next = null; //记得将尾节点的下一个指向null
        return p;
    }
}

其实翻转链表还可以使用一种递归的方式，代码如下：

ListNode * ReverseListIncur(ListNode * node, ListNode * prev)
{
    if (node == NULL)
        return NULL;
    ListNode * tmp = ReverseListIncur(node->next, node);
    node->next = prev;
    if (tmp == NULL)
        return node;
    else
        return tmp;
}