# Leetcode [No. 206] Reverse Linked List (EASY)

## 題目

https://leetcode.com/problems/reverse-linked-list/

## 思路

+ 這個就是透過while，走一圈，結束後return.

/**
 * 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)
        {
            return head;
        }
        ListNode* slow = nullptr; ListNode* fast = head->next;
        while(fast != nullptr)
        {
            head->next = slow;
            // update ptr;
            slow = head;
            head = fast;
            fast = fast->next;
        }
        head->next =slow;
        return head;
    }
};

### 解法分析
+ time complexity: O(n)

### 執行結果


---

## 用recursive版本再寫一次

+ 這個版本很重要，面試必考!!!

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (head == nullptr || head->next == nullptr) return head;
        ListNode* last = reverseList(head->next);
        head->next->next = head;
        head->next = nullptr;
        return last;
    }
};