题解 | #链表内指定区间反转#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */
//本题主要是对边界的把握，先定义好间隔节点 first,end， 然后移动M-1次。然后就需要判断一些异常情况，如果m=1呢？n等于长度呢？
public class Solution {
    /**
     * 
     * @param head ListNode类 
     * @param m int整型 
     * @param n int整型 
     * @return ListNode类
     */
    public ListNode reverseBetween (ListNode head, int m, int n) {
        // write code here
        if(m == n || head == null ||m >= n){
            return head;
        }
        
        //1.定义四个节点,分别是m位置 n 位置，还有反转前后m-1 和n+1位置节点
        ListNode first =  head, end = head;
        ListNode preFirst = null, endAfter = null;
        int step = n - m;
        //2.找到对应节点位置
        while(step > 0){
            step --;
            end = end.next;
            if(end == null){
                return head;
            }
        }
        //2.1 fist和end长度间距定义好了，真正开始遍历到m位置
        while(m - 1 > 0){
            m --;
            preFirst = first;
            first = first.next;
            end = end.next;
            if(end == null){
                return head;
            }
        }
        
        endAfter = end.next;
        end.next = null;
        reverse(first);
        if(preFirst == null){
            head = end;
        }else{
            preFirst.next = end;
        }
        first.next = endAfter;

        return head;
    }
    private void reverse(ListNode slow){
        ListNode pre = null, currentNode = slow;
        while(currentNode != null){
            ListNode next = currentNode.next;
            currentNode.next = pre;
            pre = currentNode;
            currentNode = next;
        }
        
    }
    
    
    
    
    
    
    
}