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->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.

一刷
思路：
翻转部分隔离出来，记录这部分之前的节点和之后的节点。然后翻转这部分，再和之前记录的两个节点连接起来就可以了。缺点：还是不够简洁，需要单独判断m=1的情况。希望二刷的时候能够改进。并且还要注意在某个区域内翻转时，dummy node的技巧。

Time Complexity - O(n)， Space Complexity - O(1)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if(m==n) return head;
        ListNode prev;
        if(m == 1) {
            prev = new ListNode(-1);
            prev.next = head;
        }
        else  prev = find(head, m);
         ListNode after = find(head, n+2);
         ListNode rev = reverse(prev.next, after); //number of element to get reversed
         prev.next = rev;
         if(m == 1) return prev.next;
         else return head;
    }
    
    private ListNode find(ListNode head, int m){
        while(m>2){
            head = head.next;
            m--;
        }
        return head;
    }
    
    private ListNode reverse(ListNode head, ListNode tail){
         ListNode dummy = new ListNode(-1); 
        dummy.next = tail;
        ListNode tmp = new ListNode (-1);
        while(head!=tail){
           tmp = head.next;
           head.next = dummy.next;
           dummy.next = head;
           head = tmp; 
        }
        return dummy.next;
    }
}

二刷：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
       if(head == null) return null;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        for(int i=0;i<m-1; i++) pre = pre.next;//pre.next is the start point
        
        ListNode start = pre.next;
        ListNode then = start.next;
        
        
        for(int i=0; i<n-m; i++){
            start.next = then.next;
            then.next = pre.next;
            pre.next = then;//the next to add in
            then = start.next;
        }
        
        
        return dummy.next;
    }
    
   
}