Leetcode: Reverse Linked List II

Reverse a linked list from position m to n. Do it in-place and in one-pass.For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,return 1->4->3->2->5->NULL.Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

Analysis: 这道题还是想了我蛮久，先是题意理解错了，以为只是m和n这两个位置对调，结果发现其实是m到n之间的所有元素都需要调换。一时之间没有想到怎样做reverse比较好，参考了一下网上的思路，发现这样做比较好：还是要用Runner Technique，还是要用Dummy Node；两个指针: npointer指到n的位置，mpointer指到m的前一位；每一次把mpointer后一位的元素放到npointer的后一位：mpointer.next.next = npointer.next；直到mpointer.next = npointer为止（m与n重合）

Original linked list: 1->2->3->4->5->6->7; m = 3, n =6

Step1: 1->2->4->5->6->3->7

Step2: 1->2->5->6->4->3->7

......

Result: 1->2->6->5->4->3->7

Note that pointer m is switching to right one by one in each step, but pointer n remains no change.

再次体现了在链表题中使用Dummy Node, 并且对当前node.next进行操作的好处。

Notice: 像这种链表删除插入操作，在删除插入之前，最好先拷贝一下被删除节点的下一个节点，以及插入位置的下一个节点，把它们存在一个变量ListNode store里面，直接去访问这个变量，而不要用类似mpointer.next的方式存着他们，因为这样受制于mpointer，mpointer一旦变化，就找不到这些删除/插入节点的下一个节点了。

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode reverseBetween(ListNode head, int m, int n) {
14         ListNode prev = new ListNode(-1);
15         prev.next = head;
16         ListNode mpointer = prev; //point to m-1 position
17         ListNode npointer = prev; //point to n position
18         while (m > 1) {
19             mpointer = mpointer.next;
20             m--;
21         }
22         while (n > 0) {
23             npointer = npointer.next;
24             n--;
25         }
26         while (mpointer.next != npointer) {
27             ListNode mnext = mpointer.next.next;
28             ListNode nnext = npointer.next;
29             mpointer.next.next = nnext;
30             npointer.next = mpointer.next;
31             mpointer.next = mnext;
32         }
33         return prev.next;
34     }
35 }

第二遍做法：

 1 public class Solution {
 2     public ListNode reverseBetween(ListNode head, int m, int n) {
 3         if (head == null || m > n) return head;
 4         ListNode dummy = new ListNode(-1);
 5         dummy.next = head;
 6         ListNode walker = dummy;
 7         ListNode runner = dummy;
 8         while (m > 1) {
 9             walker = walker.next;
10             m--;
11         }
12         while (n > 0) {
13             runner = runner.next;
14             n--;
15         }
16         while (walker.next != runner) {
17             ListNode cur = walker.next;
18             ListNode next = cur.next;
19             cur.next = runner.next;
20             runner.next = cur;
21             walker.next = next;
22         }
23         return dummy.next;
24     }
25 }

Leetcode: Reverse Linked List II相关推荐

LeetCode | Reverse Linked List II
这道链表反转题也是搞的我焦头烂额,好久没有写链表了,注意记忆这些 Reverse Linked List II QuestionEditorial Solution My Submissions To ...
[LeetCode] Reverse Linked List II
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1-> ...
leetcode : Reverse Linked List II [two pointers]
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1-> ...
LeetCode Reverse Linked List II
链表局部反转问题代码如下: class Solution {public ListNode reverseBetween(ListNode head, int m, int n) {if (m == ...
LeetCode Reverse Linked List II 反置链表2
题意:将指定的一段位置[m,n]的链表反置,返回链表头. 思路:主要麻烦在链表头,如果要从链表头就开始,比较特殊. 目前用DFS实现,先找到m-1的位置,再找到n+1的位置,中间这段就是否要反置的,交 ...
【??链表】LeetCode 92. Reverse Linked List II
LeetCode 92. Reverse Linked List II Solution1: 参考网址:http://www.cnblogs.com/grandyang/p/4306611.html ...
LeetCode 92. Reverse Linked List II
92. Reverse Linked List II Reverse a linked list from position m to n. Do it in one-pass. 将位置m的链接列表反 ...
(LeetCode 92)Reverse Linked List II
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1-> ...
Reverse Linked List II leetcode java
题目: Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1 ...

Leetcode: Reverse Linked List II

Leetcode: Reverse Linked List II相关推荐

最新文章

热门文章