LeetCode206:Reverse Linked List

Reverse a singly linked list.

分别用迭代和递归实现

struct ListNode
{
int val;
struct ListNode *next;
};

迭代实现：

struct ListNode* reverseList(struct ListNode* head)
{
struct ListNode *pre = NULL;
struct ListNode *cur = head;
while( cur != NULL )
{
struct ListNode *after = cur->next;
cur->next = pre;
pre = cur;
cur = after;
}
return pre;
}

递归实现：

1）如果head为空，或者只有head这一个节点，return head即可；

2）先遍历head->next为首的链表，得到一个头结点newHead；

3）把head赋值给head->next->next， head->next为空；

4）返回newHead。

struct ListNode* reverseList(struct ListNode* head)
{
if(head == NULL || head->next == NULL)
return head;

struct ListNode* newHead = reverseList(head->next);
head->next->next = head;
head->next = NULL;
return newHead;
}