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;
    }
}


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