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[LeetCode] 206. Reverse Linked List

Given the head of a singly linked list, reverse the list, and return the reversed list.

Example 1:

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]

Example 2:

Input: head = [1,2]
Output: [2,1]

Example 3:

Input: head = []
Output: []

Constraints:

The number of nodes in the list is the range [0, 5000].
-5000 <= Node.val <= 5000

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

反转链表。

这类reverse的题不会写，会写homebrew也枉然。这题我用的是iterative的思路做的。创建一个空的指针pre。当head不为空的时候，先存住head.next，然后head.next指向pre，最后pre，head，next三个指针整体往后移动一位。代码如下，也share一个注释写的比较好的discussion。

时间O(n)

空间O(1)

Java实现

 1 class Solution {
 2     public ListNode reverseList(ListNode head) {
 3         // corner case
 4         if (head == null || head.next == null) {
 5             return head;
 6         }
 7 
 8         // normal case
 9         ListNode pre = null;
10         while (head != null) {
11             ListNode next = head.next;
12             head.next = pre;
13             pre = head;
14             head = next;
15         }
16         return pre;
17     }
18 }

JavaScript实现

 1 /**
 2  * @param {ListNode} head
 3  * @return {ListNode}
 4  */
 5 var reverseList = function(head) {
 6     // corner case
 7     if (head === null || head.next === null) {
 8         return head;
 9     }
10 
11     // normal case
12     let pre = null;
13     while (head !== null) {
14         let next = head.next;
15         head.next = pre;
16         pre = head;
17         head = next;
18     }
19     return pre;
20 };