36.Reverse Linked List II-翻转链表 II（中等题）

翻转链表 II

1. 题目

   翻转链表中第m个节点到第n个节点的部分

   注意事项
   m，n满足1 ≤ m ≤ n ≤ 链表长度

2. 样例

   给出链表1->2->3->4->5->null， m = 2 和n = 4，返回1->4->3->2->5->null

3. 挑战

   在原地一次翻转完成

4. 题解

1.Stack法
对于要翻转部分的节点，用Stack辅助进行翻转，然后再与首尾部分进行拼接。

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param ListNode head is the head of the linked list 
     * @oaram m and n
     * @return: The head of the reversed ListNode
     */
    public ListNode reverseBetween(ListNode head, int m , int n) {
        if (m == n)
        {
            return head;
        }
        int count = 1;
        ListNode newHead = new ListNode(1);
        newHead.next = head;
        ListNode mPre = newHead;
        ListNode nTail = null;
        Stack<ListNode> stack = new Stack<>();
        while (head != null)
        {
            if (count < m)
            {
                mPre = mPre.next;
            }
            else if (count >= m && count <= n)
            {
                stack.push(head);
                if (count == n)
                {
                    nTail = head.next;
                    break;
                }
            }
            head = head.next;
            count++;
        }
        while (!stack.isEmpty())
        {
            mPre.next = stack.pop();
            mPre = mPre.next;
        }
        mPre.next = nTail;
        return newHead.next;
    }
}

2.原地翻转
为了方便处理，先给链表加上一个头结点dummy，dummy->1->2->3->4->5->null。然后将head指向dummy节点，开始遍历链表，将head移动到第一个需要翻转的节点的前节点，即head指向1。
先保存3的后继节点，再将节点3指向其前节点2，变成dummy->1->2<-3 4->5->null，再将节点后移一位，继续处理节点3和节点4,。最后将节点首尾相连即可。

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param ListNode head is the head of the linked list 
     * @oaram m and n
     * @return: The head of the reversed ListNode
     */
    public ListNode reverseBetween(ListNode head, int m , int n) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        for (int i = 1; i < m; i++) 
        {
            head = head.next;
        }

        ListNode mNode = head.next;
        ListNode nNode = mNode;
        ListNode postnNode = mNode.next;
        for (int i = m; i < n; i++) 
        {
            ListNode temp = postnNode.next;
            postnNode.next = mNode;
            mNode = postnNode;
            postnNode = temp;
        }
        nNode.next = postnNode;
        head.next = mNode;

        return dummy.next;
    }
}

Last Update 2016.9.26