LeetCode 25. Reverse Nodes in k-Group

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本文介绍了一种链表操作的方法：将链表按k个一组进行节点的反转。通过迭代方式实现，确保只使用常数级额外空间，并保持原有节点值不变。文章提供了完整的Java代码实现。

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
k is a positive integer and is less than or equal to the length of the linked 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

题目描述
这是leetcode上一个hard的题目。将链表分为k组，每一组进行反转。
关键是要找好前一个链表的尾结点，新链表的头结点和尾结点。


public class ReverseNodesInKGroup {
    public ListNode reverseKGroup(ListNode head, int k) {
        if ( k==1 || k==0 || head==null || head.next==null )
            return head; 
        else {
            ListNode dummy = new ListNode(-1);
            dummy.next = head; 
            ListNode cur = head, //新链表的尾结点
                    op = head.next, //即将操作的结点
                    list = head, //新链表的头结点
                    pre = dummy, //上一个链表的尾结点
                    temp = head; 
            // 获取链表的长度
            int len = 0 ; 
            while ( temp!=null ) {
                ++len; 
                temp = temp.next;
            }
            for ( int i=0 ;i+k<=len; i+=k ) {
                for ( int j=1; j<k; j++ ) {
                    cur.next = op.next; //!!
                    op.next = list; 
                    list = op; 
                    op = cur.next;
                }
                pre.next = list;
                pre = cur;
                cur = op;
                if ( cur==null ) 
                    break; 
                op = cur.next;
                list = cur; 
            }

            return dummy.next; 
        }
    }
}