题解 | #链表内指定区间反转#

import java.util.*;
/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */
public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     * @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) {
        int i=1;
        ListNode currNode = head;
        ListNode pre = head;
        ListNode mn_head = null;
        while(currNode!=null){
            if(i>=m && i<=n){
                ListNode afterMNode = currNode; //第n+1个节点
                for (int j=0;j<n-m+1;j++){
                    afterMNode = afterMNode.next;
                }
                mn_head = reverseMN(currNode,n - m + 1);
                if(i==1){
                    head = mn_head;
                }else pre.next = mn_head;
                currNode.next = afterMNode;
                break;
            }else {
                pre = currNode;
                currNode = currNode.next;
                i++;
            }
        }
        return head;
    }

    public ListNode reverseMN(ListNode currNode,int n){
        if (n==1){
            return currNode;
        }
        ListNode new_head = reverseMN(currNode.next,--n);
        currNode.next.next = currNode;
        currNode.next = null;
        return new_head;
    }

}

分三段处理   0~m-1   m~n   n+1~最后