92. Reverse Linked List II

题目：

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.

链接： http://leetcode.com/problems/reverse-linked-list-ii/

6/4/2017

注意

1. m和n的循环判断条件不一样，m是先－再判断，n是先判断再－

2. 很多断点处需要额外用ListNode varaible来标记，比如，prev, second, dummy2

 1 public class Solution {
 2     public ListNode reverseBetween(ListNode head, int m, int n) {
 3         ListNode dummy = new ListNode(-1);
 4         dummy.next = head;
 5         ListNode prev = dummy;
 6 
 7         n -= m;
 8         while (--m > 0) {
 9             prev = prev.next;
10         }
11         ListNode cur = prev.next;
12         prev.next = null;
13         ListNode dummy2 = new ListNode(-1);
14         ListNode second = cur;
15 
16         while (n >= 0) {
17             ListNode next = cur.next;
18             cur.next = dummy2.next;
19             dummy2.next = cur;
20             cur = next;
21             n--;
22         }
23         second.next = cur;
24         prev.next = dummy2.next;
25         return dummy.next;
26     }
27 }

别人的答案

非常清晰的写法

学习的地方：

1. 第6行用for循环不是while循环

2. 第二个循环里，start是不变的，start.next一直在变，then总是指向当前结束时start.next，prev.next是当前的then

https://discuss.leetcode.com/topic/8976/simple-java-solution-with-clear-explanation

 1 public ListNode reverseBetween(ListNode head, int m, int n) {
 2     if(head == null) return null;
 3     ListNode dummy = new ListNode(0); // create a dummy node to mark the head of this list
 4     dummy.next = head;
 5     ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing
 6     for(int i = 0; i<m-1; i++) pre = pre.next;
 7     
 8     ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversed
 9     ListNode then = start.next; // a pointer to a node that will be reversed
10     
11     // 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3
12     // dummy-> 1 -> 2 -> 3 -> 4 -> 5
13     
14     for(int i=0; i<n-m; i++)
15     {
16         start.next = then.next;
17         then.next = pre.next;
18         pre.next = then;
19         then = start.next;
20     }
21     
22     // first reversing : dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4
23     // second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish)
24     
25     return dummy.next;
26     
27 }