Reverse Nodes in k-Group k-group翻转链表@LeetCode

这种题目是要背下来的。。

翻转单链表：

ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode pre = dummy;
    ListNode cur = head.next;
    ListNode last = head;
    while(cur != null){
        last.next = cur.next;
        cur.next = pre.next;
        pre.next = cur;
        cur = last.next;
    }
    head = dummy.next;

package Level3;

import Utility.ListNode;

/**
 * Reverse Nodes in k-Group
 *
 *Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5
 */
public class S25 {

	public static void main(String[] args) {

	}
	
	public ListNode reverseKGroup(ListNode head, int k) {
        if(head==null || k==1){
        	return head;
        }
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        int i = 0;
        while(head != null){
        	i++;
        	if(i%k == 0){
        		pre = reverse(pre, head.next);
        		head = pre.next;
        	}else{
        		head = head.next;
        	}
        }
        return dummy.next;
    }
	
	
	/**
     * Reverse a link list between pre and next exclusively
     * an example:
     * a linked list:
     * 0->1->2->3->4->5->6
     * |           |   
     * pre        next
     * after call pre = reverse(pre, next)
     * 
     * 0->3->2->1->4->5->6
     *          |  |
     *          pre next
     * @param pre 
     * @param next
     * @return the reversed list's last node, which is the precedence of parameter next
     */
    private static ListNode reverse(ListNode pre, ListNode next){
        ListNode last = pre.next;//where first will be doomed "last"
        ListNode cur = last.next;
        while(cur != next){
            last.next = cur.next;
            cur.next = pre.next;
            pre.next = cur;
            cur = last.next;
        }
        return last;
    }

}

Ref:  http://www.cnblogs.com/lichen782/p/leetcode_Reverse_Nodes_in_kGroup.html