LeetCode-Reverse Nodes in k-Group

直接数了个数 算了要做几个reverse 然后直接做的

可以优化的就是不需要数有多少个 每次累加到够了k就做一次 其实差不多

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode reverseKGroup(ListNode head, int k) {        if ( head == null || k <= 1 )            return head;        int length = 0;        ListNode pt = head;        while ( pt != null ){            pt = pt.next;            length ++;        }        int times = length / k;        ListNode dummy = new ListNode ( 0 );        dummy.next = head;        pt = dummy;        for ( int i = 0; i < times; i ++ ){            ListNode pre = head, cur = head.next;            for ( int j = 1; j < k; j ++ ){                ListNode next = cur.next;                cur.next = pre;                pre = cur;                cur = next;            }            pt.next.next = cur;            pt.next = pre;            pt = head;            head = head.next;        }        return dummy.next;    }}