[Leetcode] Reverse Linked List

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题目链接在此

Reverse a singly linked list.

反转一个单链表。别小看这道题啊，我TX面试上来就问这道题。当时我觉得用两个指针就够了——显然天真了，得用三个指针。

这是一个循环的算法：

struct ListNode {    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL) {}};class Solution {public:ListNode* reverseList(ListNode* head) {if (head == NULL || head->next == NULL)return head;ListNode *p = head;ListNode *q = head->next;ListNode *r = head->next->next;head->next = NULL;while (true) {q->next = p;p = q;q = r;if (r != NULL)r = r->next;elsebreak;};return p;}};

以下则是递归做法：

class Solution2 {public:ListNode* reverseList(ListNode* head) {if (head == NULL || head->next == NULL)return head;ListNode* p = head;while (p->next != NULL)p = p->next;reverse(head);return p;}private:ListNode* reverse(ListNode* head) {if (head->next == NULL)return head;else {ListNode* tail = reverse(head->next);tail->next = head;tail = tail->next;tail->next = NULL;return tail;}}};

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