92. Reverse Linked List II

Problem:

Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

5/1/2017 update:
第三遍是自己想出来的one pass。
主要是搞清楚怎么把m 到n reverse过后的头和尾和原链表链接起来。

class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy;
        ListNode beforeM = null;
        
        for (int i = 0; i < m - 1; i++) {
            prev = prev.next;
        }

        beforeM = prev;
        prev = prev.next;
        
        ListNode cur = prev.next, tail = prev;
        for (int i = 0; i < n - m; i++) {
            ListNode next = cur.next;
            cur.next = prev;
            prev = cur;
            cur = next;
        }
        beforeM.next = prev;
        tail.next = cur;
        return dummy.next;
    }
}

12/21/2017 update
The last step is the most import part of this problem. Don't forget it.

-------------------------------------------------------------------------------
思路：
找到m-1节点，保存下来，相当于子链表的dummy。然后从m开始，reverse n-m次。初始化prev在m上，cur 在m.next上。reverse结束后，prev在n上，cur在n.next上。最后重新连接链表。
做了这道题和reserve nodes in k-group才真正掌握了dummy node相关题目。
reverse过程如下图：


实现：
九章的解法太复杂了。需要的临时节点太多容易出错。代码如下：

 
 public class Solution {   
   public ListNode reverseBetween(ListNode head, int m, int n) {  
     // write your code here  
     ListNode dummy = new ListNode(0);  
     dummy.next = head;  
     head = dummy;  
     for(int i = 0; i < m-1; i++) {  
       head = head.next;  
     }  
     ListNode prev = head.next;  
     ListNode cur = prev.next;  
     for(int i = 0; i < n-m; i++) {  
       ListNode temp = cur.next;  
       cur.next = prev;  
       prev = cur;  
       cur = temp;  
     }  
     head.next.next = cur;  
     head.next = prev;  
     return dummy.next;  
   }  
 }