LeetCode刷题笔录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.

想法是找到m到n的这一段链表，做一个通常的reverse，然后把reverse的结果接回到原来的链表中间。注意一些边界条件。我们一开始要找到m的前一个节点，用来将reverse之后的链表接回去。但如果m=1,即m=head节点，那么这个beforeM节点就没用了。

/**
 * 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) {
        boolean isFromStart = false;
        if(m == 1){
            isFromStart = true;
        }
        ListNode beforeM = head;
        for(int i = 1; i < m - 1; i++){
            beforeM = beforeM.next;
        }
       
        ListNode afterN = head;
        ListNode N = head;
        for(int i = 1; i <= n; i++){
            afterN = afterN.next;
            if(i < n){
                N = N.next;
            }
        }
        //unhook N from the list
        N.next = null;
        //reverse between m and n
        ListNode newM = reverse(isFromStart ? head : beforeM.next);
        ListNode newN = newM;
        while(newN.next != null){
            newN = newN.next;
        }
        //link the reversed list back to the original
        if(!isFromStart){
            beforeM.next = newM;
        }
        newN.next = afterN;
        
        if(isFromStart)
            return newM;
        else
            return head;
        
    }
    
    public ListNode reverse(ListNode head){
        ListNode prev = null;
        ListNode cur = head;
        ListNode next;
        while(cur != null){
            next = cur.next;
            cur.next = prev;
            prev = cur;
            cur = next;
        }
        return prev;
    }
}