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[LeetCode] 92. Reverse Linked List II

Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list.

Example 1:

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

Example 2:

Input: head = [5], left = 1, right = 1
Output: [5]

Constraints:

The number of nodes in the list is n.
1 <= n <= 500
-500 <= Node.val <= 500
1 <= left <= right <= n

Follow up: Could you do it in one pass?

反转链表II。

也是不会做，会写homebrew也枉然的题目，但是非常容易错。题意是给一个链表和两个数字m和n，请反转m和n之间的node，只能走一次。

这题我借助了一个图，非得看着图做才行。

/*

1 -> 2 -> 3 -> 4 -> 5

guard p next

*/

思路是头插法。我参考了一个很好的图示。照着例子跑一下吧。假设需要反转的部分是从2到4，反转的顺序是先把3拿出来插入1和2中间，然后再把4插入1和3中间，像这样。guard永远是需要反转的部分之前的一个节点，像个守卫一样不能动！p是需要反转的部分的第一个节点。

1 -> 2 -> 3 -> 4 -> 5

1 -> 3 -> 2 -> 4 -> 5

1 -> 4 -> 3 -> 2 -> 5

时间O(n)

空间O(1)

JavaScript实现

 1 /**
 2  * @param {ListNode} head
 3  * @param {number} m
 4  * @param {number} n
 5  * @return {ListNode}
 6  */
 7 var reverseBetween = function(head, m, n) {
 8     let dummy = new ListNode(0);
 9     dummy.next = head;
10     let pre = dummy;
11     let cur = dummy.next;
12     for (let i = 1; i < m; i++) {
13         cur = cur.next;
14         pre = pre.next;
15     }
16     for (let i = 0; i < n - m; i++) {
17         let temp = cur.next;
18         cur.next = temp.next;
19         temp.next = pre.next;
20         pre.next = temp;
21     }
22     return dummy.next;
23 };

Java实现

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        // corner case
        if (head == null) {
            return null;
        }

        // normal case
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        for (int i = 1; i < m; i++) {
            pre = pre.next;
        }
        ListNode cur = pre.next;
        for (int i = 0; i < n - m; i++) {
            ListNode next = cur.next;
            cur.next = next.next;
            next.next = pre.next;
            pre.next = next;
        }
        return dummy.next;
    }
}