92. Reverse Linked List II

题目：

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.

链接： http://leetcode.com/problems/reverse-linked-list-ii/

题解：

把翻转部分隔离出来，记录这部分之前的节点和之后的节点。然后翻转这部分，再和之前记录的两个节点连接起来就可以了。

Time Complexity - O(n)， Space Complexity - O(1)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if(head == null || head.next == null || m == n)
            return head;
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode preHeadToReverse = dummy, tailToReverse = dummy;
        int count = 0;
        
        while(count < n) {
            if(tailToReverse == null)
                return dummy.next;
            if(count < m - 1)
                preHeadToReverse = preHeadToReverse.next;
            tailToReverse = tailToReverse.next;
            count++;
        }
        
        ListNode headToReverse = preHeadToReverse.next, postTailToReverse = tailToReverse.next;
        tailToReverse.next = null;
        preHeadToReverse.next = reverse(headToReverse);
        headToReverse.next = postTailToReverse;
        return dummy.next;
    }
    
    private ListNode reverse(ListNode head) {
        if(head == null || head.next == null)
            return head;
        ListNode dummy = new ListNode(-1);
        
        while(head != null) {
            ListNode tmp = head.next;
            head.next = dummy.next;
            dummy.next = head;
            head = tmp;
        }
        
        return dummy.next;
    }
}

这道题目应该和其他几道联合起来做，比如按照先后顺序完成以下几道题目

1) Reverse Linked List

2) Reverse Linked List II

3) Swap Node in Pairs

4) Swap Node in K-Gruo

二刷：

我们需要记录四个变量， pre， headToReverse，tailToReverse， postTail，然后当遍历时节点在headToReverse以及tailToReverse的时候我们反转。下面我写得比较麻烦，单独把reverse写成了一个函数，而切要先找到结尾点才能reverse，这样就多遍历了一遍反转区域，还需要好好改写。

Java:

Time Complexity - O(n)， Space Complexity - O(1)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (head == null || head.next == null || m == n) {
            return head;
        }
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode preHeadToReverse = dummy, tailToReverse = dummy;
        while (m > 1 || n > 0) {
            if (m > 1) {
                preHeadToReverse = preHeadToReverse.next;
                m--;
            }
            if (n > 0) {
                tailToReverse = tailToReverse.next;
                n--;
            }
        }
        ListNode headToReverse = preHeadToReverse.next, postTailToReverse = tailToReverse.next;
        tailToReverse.next = null;
        preHeadToReverse.next = reverse(headToReverse); 
        headToReverse.next = postTailToReverse;
        return dummy.next;
    }
    
    private ListNode reverse(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode dummy = new ListNode(-1); 
        ListNode tmp = new ListNode (-1);
        while (head != null) {
            tmp = head.next;
            head.next = dummy.next;
            dummy.next = head;
            head = tmp;
        }
        return dummy.next;
    }
}

下面直接one pass。我们先设置fakeHead dummy，让preHead = dummy，找的过程中当m > 1的时候 preHeadToReverse = preHeadToReverse .next，然后找到headToReverse = preHeadToReverse.next，也设置一个标记tail = headToReverse，最后用来添加连接反转前节点n的后一个节点postTail。

接下来我们先设置preHeadToReverse.next = null，在(n > 0以及headToReverse != null)的情况下，我们利用reverse linkedlist的方法遍历这部分反转区域。遍历完毕以后的headToReverse就等于反转前节点n的下一个节点postTail，这时候我们只需要将两部分连接起来，做一个 tail.next = headToReverse就可以了。还能再简化。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (head == null || head.next == null || m == n) {
            return head;
        }
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode preHeadToReverse = dummy;
        while (m > 1) {
            preHeadToReverse = preHeadToReverse.next;
            m--;
            n--;
        }
        ListNode headToReverse = preHeadToReverse.next;
        ListNode tail = headToReverse, node = headToReverse;
        preHeadToReverse.next = null;
        while (n > 0 && headToReverse != null) {
            node = headToReverse.next;
            headToReverse.next = preHeadToReverse.next;
            preHeadToReverse.next = headToReverse;
            headToReverse = node;
            n--;
        }
        tail.next = headToReverse;
        return dummy.next;
    }
}

三刷:

方法和二刷一样。

先找到m的前一个节点pre，这里要注意是在m > 1的条件下进行遍历
设置head = pre.next，用来进行遍历，设置tail = head，因为翻转完毕以后这个head节点应该处于最后，所以我们设置这个tail用来连接n后面的节点们
设置tmp = head，也可以直接设置tmp为null，用来保存时head下一位置的节点
设置pre.next = null
在n > 0的情况下进行翻转，每次n--
最后连接tail和tmp，然后返回结果

Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode pre = dummy;
        
        while (m > 1) {     
            pre = pre.next;
            m--;
            n--;
        }
        
        head = pre.next;
        ListNode tail = head;
        ListNode tmp = head;
        pre.next = null;
        
        while (n > 0) {
            tmp = head.next;
            head.next = pre.next;
            pre.next = head;
            head = tmp;
            n--;
        }
        
        tail.next = tmp;
        return dummy.next;
    }
}

Reference:

https://leetcode.com/discuss/10794/share-my-java-code

https://leetcode.com/discuss/25580/simple-java-solution-with-clear-explanation

https://leetcode.com/discuss/35440/240ms-java-solution

https://leetcode.com/discuss/72660/short-java-solution-for-reverse-linked-list-ii