[Leetcode] 206. Reverse Linked List 解题报告

题目：

Reverse a singly linked list.

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Hint:

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

思路：

1、递归法：如果头结点为空或者链表中只有一个结点，则属于平凡情况，直接返回。否则首先翻转以head->next为头结点的链表，然后把head接在末尾，并将head的next置为空。

2、迭代法：从前往后扫描链表，每次翻转一下next的指向，直到到达链表末尾。最后别忘了将head的next置为空。

代码：

1、递归法：

/** * 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 || head->next == NULL) {            return head;        }        ListNode *new_tail = head->next;        ListNode *new_head = reverseList(head->next);        new_tail->next = head;        head->next = NULL;        return new_head;    }};

2、迭代法：

/** * 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;        }        ListNode* pre_node = head;        ListNode* crt_node = head->next;        while(crt_node != NULL)        {            ListNode* nxt_node = crt_node->next;    // backup            crt_node->next = pre_node;            pre_node = crt_node;            crt_node = nxt_node;        }        head->next = NULL;        return pre_node;    }};