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) {
        // write code here
        if(head.next == null){
            return head;
        }

        if(m == n){
            return head;
        }
        
        Stack<ListNode> stack = new Stack<>();
        ListNode originHead = head;
        ListNode mNode;
        ListNode nNode;
        //m节点的前一个节点
        ListNode premNode = null;
        //n节点的后一个节点
        ListNode afternNode = null;

        //把mNode到nNode这些节点放到stack中
        for(int i = 1; i<m; i++){
            if(i == m-1){
                premNode = head;
            }
            head = head.next;
        }
        mNode = head;
        stack.add(mNode);
        for(int j=0; j<n-m; j++){
            head = head.next;
            stack.add(head);
        }
        nNode = head;
        afternNode = nNode.next;

        //对stack中的元素反转
        ListNode node = stack.pop();
        //区间内链表头结点
        ListNode temHead = node;
        do{
            ListNode temNode = stack.pop();
            node.next = temNode;
            node = temNode;
        }while(!stack.isEmpty());

        //如果m==1，temHead做为最终列表的头节点
        if(m == 1){
            originHead = temHead;
        }else{
            premNode.next = temHead;
        }
        
        node.next = afternNode;
        return originHead;
    }
}