Reverse a singly linked list.
常规方法，遍历一遍，迭代计算。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* reverseList(struct ListNode* head) 
{
    if(NULL == head)
        return NULL;
    
    struct ListNode * p = head->next;
    /*第一个结点的next置为NULL*/
    head->next = NULL;
    /*中间变量，用于保存当前处理节点的下一个节点*/
    /*处理结束后赋值给p，开始下一次循环*/
    struct ListNode *ptmp = NULL;
    while(p)
    {
        /*保存当前处理结点的下一个节点*/
        ptmp = p->next;
        p->next = head;
        head = p;
        /*当前处理结点的下一个节点赋值给p*/
        p = ptmp;
        
    }
    
    return head; 
}

递归实现,gdb跟了下，自己知道大致思路，却总AC不了，参考大神的Java代码，编写如下递归形式的代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* reverseList(struct ListNode* head) 
{
    if(NULL==head||NULL==head->next) 
        return head;

    struct ListNode *p=head->next;
    
    head->next=NULL;
    
    struct ListNode *newhead = reverseList(p);
    
    p->next=head;

    return newhead;
}

另一种相对清晰的解法：

/**
 * 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) {
        ListNode *curr = head,*prev = NULL;
        while(curr != NULL)
        {
            ListNode *temp = curr->next;
            curr->next = prev;
            prev = curr;
            curr = temp;
        }
        return prev;
    }
};

较好理解的递归实现：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(!head || !head->next)
            return head;
        ListNode* node = reverseList(head->next);
        head->next->next = head;
        head->next = nullptr;

        return node;
    }
};

C++迭代实现

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* prev = nullptr,*curr = head;
        while(curr)
        {
            ListNode* tmp = curr->next;
            curr->next = prev;
            prev = curr;
            curr = tmp;
        }

        return prev;
    }
};