Reverse Nodes in k-Group @leetcode

ReverseKGroup.java

package leetcode.linkedlist;
/**
 * Solution: 拆分group, 每个group上reverse, 思路并不难，但是实现起来挺麻烦，
 * 每一段需要currentHead, currentTail 做标记用于后面的链接. 还有preTail. 另外reverse过程中需要preNode和nextNode
 * 有时候忘记这些flag就不容易一下子写出来，所以要按功能记，reverse需要2个，连接时候需要3个.
 * @author jeffwan
 * @date Apr 15, 2014
 */
public class ReverseKGroup {	
	public ListNode reverseKGroup(ListNode head, int k) {
		if (head == null || k <= 1) {
			return head;
		}

		int length = getLength(head);
		int group = length / k;
		if (group == 0) {
			return head;
		}
	
		ListNode current = head;
		ListNode preTail = null, currentHead = null, currentTail = null;
		ListNode preNode = null, nextNode = null;
		
		for (int i = 0; i < group; i++) {
			preNode = null;
			for (int j = 0; j < k; j++) {
				// Reverse
				if (current != null) {
					nextNode = current.next;
					current.next = preNode;
					preNode = current;
				}
				// Assign currentHead, currentTail for future concatenate.
				if (j == 0) {
					currentTail = current;
				}
				if (j == k - 1) {
					currentHead = current;
				}
				current = nextNode;
			}
			
			if (preTail == null) {
				head = currentHead;
			} else {
				preTail.next = currentHead;
			}
			
			preTail = currentTail;
		}
		
		currentTail.next = current;
		return head;
	}
	
	public int getLength(ListNode head) {
		int length = 0;
		while (head != null) {
			head = head.next;
			length++;
		}
		
		return length;
	}

	class ListNode {
		int val;
		ListNode next;
		ListNode (int x) { this.val = x; }
	}
}