Leetcode - Reverse Nodes in k-Group

[分析]
使用递归会Memory Limit Exceed，迭代方式参考
https://leetcode.com/discuss/17483/share-my-java-solution-with-comments-in-line
解析见代码注释
Reverse Linked List， Reverse Linked List II 可使用此题的解题技巧，可联系着体会，而Reorder List中包含Reverse Linked List步骤。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    // prev->head->...->...->tail->|...| //init state of reversing k group [head, tail]
    // prev->tmp->...->...->tail->|...|
    // prev->...->...->tail->tmp->|...| //delete tmp & insert at tail.next
    // prev->...->tail->tmp'->...->head->|...| //keep doing until prev.next = tail
    // prev->tail->...->...->head->|...| // finish state
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null || head.next == null || k < 2)
            return head;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy, tail = head, tmp;
        while (true) {
            // check remain length of list
            int count = 0;
            while (tail != null) {
                count++;
                if (count == k) break;
                tail = tail.next;
            }
            if (tail == null) break;
            // reverse a K group
            head = prev.next;
            while (prev.next != tail) {
                tmp = prev.next;
                prev.next = tmp.next; // delete tmp from prev.next
                tmp.next = tail.next; // step 1 of inserting tmp at tail.next
                tail.next = tmp; // step 2 of inserting tmp at tail.next
            }
            prev = head;
            tail = head.next;
        }
        return dummy.next;
    }
}

public class Solution {
    public ListNode reverseList(ListNode head) {
        if (head == null || head.next == null)
            return head;
        ListNode curr = head.next;
        head.next = null;
        ListNode prev = head, next;
        while (curr != null) {
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }
    public ListNode reverseList1(ListNode head) {
        if (head == null || head.next == null)
            return head;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode tail = head;
        while (tail.next != null) {
            tail = tail.next;
        }
        ListNode tmp;
        while (dummy.next != tail) {
            tmp = dummy.next;
            dummy.next = tmp.next;
            tmp.next = tail.next;
            tail.next = tmp;
        }
        return tail;
    }
}

public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        int count = 1;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy; //prev is the (m-1)th node
        while (count < m) {
            prev = prev.next; 
            count++;
        }
        ListNode tail = prev.next;
        while (count < n) { // tail is the nth node
            tail = tail.next;
            count++;
        }
        ListNode tmp;
        while (prev.next != tail) {
            tmp = prev.next;
            prev.next = tmp.next;
            tmp.next = tail.next;
            tail.next = tmp;
        }
        return dummy.next;
    }
}