Reverse Linked List II

Reverse Linked List II (Upgrade Version for Rotate List)

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.


Naive Way:一开始想了一个不断交换的方法，这样可以做到in-place,但是交换之后最大的麻烦就是无法在O(1)的时间找回远端的节点的父节点，所以这个方法不符合O(n)的复杂度。然后才想到了这个不断插入的做法。现将远端的指针和近端指针摆好，不断将近端指针指向的插入远端指针后面，直到近端指向远端指针。也可以不断将近端后面的指针插入近端前面，知道近端和远端相遇。

算法复杂度是O(n), space O(1)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        ListNode newHead = new ListNode(0);
        newHead.next = head;
        ListNode p1 = newHead, p2 = p1;
        int i = 0;
        while(++i <= n && p2.next!=null){
            if(i < m) p1 = p1.next;
            p2 = p2.next;
        }
        // since n <= length of list, no need to check i<n
        while(p1.next!=p2){
            ListNode temp = p1.next;
            p1.next = temp.next;
            
            temp.next = p2.next;
            p2.next = temp;
        }
        return newHead.next;
    }
}