206 反转链表

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

进阶: 你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

解法

创建两个指针cur，pre，让cur->next指向pre，pre指向cur，cur指向cur->next。

• 迭代(Python)

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        pre, cur = None, head
        while cur:
            tmp = cur.next
            cur.next = pre
            pre = cur 
            cur = tmp
        return pre

• 迭代(C++)

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

• 递归

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if head == None or head.next == None:
            return head
        newNode = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return newNode