LeetCode 206. Reverse Linked List(C++)

题目：

Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

分析：

分别用迭代和递归来实现。

迭代就是新建一个newhead节点，遍历原链表，将每一个node接到newhead，注意保存head->next用来更新head。

递归则是利用函数先走到链表尾端，依次更新每一个node的next，最后返回newhead。

程序：

//iteratively
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode *newhead = NULL;
        while(head){
            ListNode *p = head->next;
            head->next = newhead;
            newhead = head;
            head = p;
        }
        return newhead;
    }
};

//recursively
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (head == NULL || head->next == NULL){
            return head;
        }
        ListNode* newhead = reverseList(head->next);
        head->next->next = head;
        head->next = NULL;
        return newhead;
    }
};

原文地址：https://www.cnblogs.com/silentteller/p/10459455.html