Reverse a singly linked list.
A linked list can be reversed either iteratively or recursively. Could you implement both?
翻转链表,非常经典的一道题目,不难,但是细节问题需要注意,这里用三种方法来做,前两种是迭代,最后一种是递归,个人认为第一种方法最好。
方法一:利用pre指针,追个翻转节点
该方法从头到尾遍历链表,逐个翻转节点就行,易错点在于返回值一定是pre,而且pre的初始化要是null,这里就是反转之后链表的末尾,第一次我就写成了cur
代码如下:
1 /**
2 * Definition for singly-linked list.
3 * struct ListNode {
4 * int val;
5 * ListNode *next;
6 * ListNode(int x) : val(x), next(NULL) {}
7 * };
8 */
9 class Solution {
10 public:
11 ListNode* reverseList(ListNode* head) {
12 if (!head || !head->next)
13 return head;
14 ListNode *pre = nullptr, *cur = head; //初始化一样很重要
15 while (cur)
16 {
17 ListNode *t = cur->next;
18 cur->next = pre;
19 pre = cur;
20 cur = t;
21 }
22 //return cur;
23 return pre;
24 }
25 };
时间复杂度:O(n)
空间复杂度:O(1)
方法二:
设置虚拟头结点,每次在虚拟节点之后插入cur节点,最后直接返回dummy->next
1 /**
2 * Definition for singly-linked list.
3 * struct ListNode {
4 * int val;
5 * ListNode *next;
6 * ListNode(int x) : val(x), next(NULL) {}
7 * };
8 */
9 class Solution {
10 public:
11 ListNode* reverseList(ListNode* head) {
12 if (!head || !head->next)
13 return head;
14 ListNode *dummy = new ListNode(-1);
15 ListNode *pre = dummy;
16 ListNode *cur = head;
17 dummy->next = head;
18 while (cur && cur->next)
19 {
20 ListNode *t = cur->next;
21 cur->next = t->next;
22 t->next = pre->next; //这里注意 一定不要写成t->next = cur;这么做只有第一次交换是对的之后都不对
23 pre->next = t;
24 }
25 return dummy->next;
26 }
27 };
时间复杂度:O(n)
空间复杂度:O(1)
方法三:递归的方法,每次翻转当前节点之后的所有节点,注意递归之后的操作以及最后的终止条件的确定
1 /**
2 * Definition for singly-linked list.
3 * struct ListNode {
4 * int val;
5 * ListNode *next;
6 * ListNode(int x) : val(x), next(NULL) {}
7 * };
8 */
9 class Solution {
10 public:
11 ListNode* reverseList(ListNode* head) {
12 if (!head || !head->next)
13 return head;
14 ListNode *newHead = reverseList(head->next);
15 head->next->next = head;
16 head->next = nullptr;
17 return newHead;
18 }
19 };
时间复杂度:O(n)
空间复杂度:O(n)
参考连接:
https://leetcode.com/problems/reverse-linked-list/solution/
https://leetcode.com/problems/reverse-linked-list/discuss/58130