【LeetCode】206. Reverse Linked List

题目：

Reverse a singly linked list.

提示：

此题不难，可以用迭代或者递归两种方法求解。记得要把原来的链表头的next置为NULL；

代码：

迭代：

/**
 * 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) return NULL;
        ListNode *pre = head, *cur = head->next;
        pre->next = NULL;
        ListNode *tmp;
        while (cur) {
            tmp = cur->next;
            cur->next = pre;
            pre = cur;
            cur = tmp;
        }
        return pre;
    }
};

递归：

/**
 * 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) {
        return reverseNode(head, NULL);
    }

    ListNode* reverseNode(ListNode* head, ListNode* newHead) {
        if (!head) return newHead;
        ListNode *next = head->next;
        head->next = newHead;
        return reverseNode(next, head);
    }
};