LeetCode-Reverse Linked List

继续一道链表的算法题，运用笨笨的方法实现了（不过却是8ms，不可思议）

/**
* 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;
else if(head->next == NULL) return head;
else if(head->next->next == NULL)
{
head->next->next = head;
head = head->next;
head->next->next = NULL;
return head;
}
else
{
ListNode *a, *b, *c, *pre, *current;
pre = head->next;
current = pre->next;
a = pre;
b = current;
c = current->next;
pre->next = head;
head->next = NULL;
while(current->next != NULL)
{

c = current->next;
current->next = a;
a = b;
b = c;
current = c;
}
current->next = a;
head = b;
return head;
}
}
};

然后找大神的博客看，发现了别人家的代码

（1）非递归版

//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;
}
};

此外还发现了用递归实现的更强版本（递归思想：层层进入最后一个节点，然后把前节点的指针付给后节点的指针）

（2）递归版本

//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;
}
};

本题用的递归实现要求记住