LeetCode 92. Reverse Linked List II

原题链接在这里：https://leetcode.com/problems/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.

题解：

本题是Reverse Linked List的拓展。

基本思路类似Reverse Linked List的iteration.

需先找到reverse范围的前一个点记为mark. 翻转时while loop走一次，翻转一个点，所以用一个计数器i 来限定接下来翻转的次数知道 i<n.

翻转完得到的的cur是翻转后的表头，把它连载mark的后面.

Note: 当m==length时，mark.next == null 需单独讨论，否则下面tail = mark.next, 直接用tail.next会NRE.

Time Complexity: O(n), one pass.

Space: O(1).

AC  Java: 

/**
 * 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 || head.next == null){
            return head;
        }
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode mark = dummy;
        int i = 1;
        while(i<m){
            mark = mark.next;
            i++;
        }
        
        if(mark.next == null){  //tail = mark.next, tail.next will throw NRE without this check
            return dummy.next;
        }
        
        ListNode tail = mark.next;
        ListNode cur = mark.next;
        ListNode pre;
        ListNode temp;
        while(tail.next != null && i<n){
            pre = cur;
            cur = tail.next;
            temp = cur.next;
            cur.next = pre;
            tail.next = temp;
            i++;
        }
        mark.next = cur;
        return dummy.next;
    }
}