leetcode: reverse-nodes-in-k-group

题目描述:

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

解题思路：

1. 这道题的意思就是说，给定参数 k,每次反转链表中的 k 个节点，不足 k 个就不操作
2. 这道题考察的核心还是反转，所以思路基本上一样，唯一的不同就是先算出来链表的长度除以 k ,得到需要反转的子链条数
3. 这道题隐形的好处就是不用再去考虑边界条件

代码如下：

    public ListNode reverseKGroup(ListNode head, int k) {

        ListNode dummy = new ListNode(-1);
        dummy.next = head;

        ListNode start = head;
	
	// 求出链表的长度
        int count = 0;
        while(start != null){
            count++;
            start = start.next;
        }

	// 求出子链表的条数
        start = dummy;
        count = count / k;

	// 双层循环，分布反转 
        for(int i = 0; i < count; i++){
            ListNode p = start.next;
            for(int j = 1; j < k; j++){
                ListNode temp = p.next;
                p.next = temp.next;
                temp.next = start.next;
                start.next = temp;
            }
            start = p;
        }
        return dummy.next;
    }

转载于:https://my.oschina.net/happywe/blog/3084933