题解 | #牛群旋转#

考察双指针模拟。找到倒数第k个位置将链表断开重连就可以了。

完整Java代码如下所示

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    public ListNode rotateLeft (ListNode head, int k) {
        // write code here
        ListNode temp = head;
        int length = 0;
        while(temp!=null){
            length++;
            temp = temp.next;
        }
        k = k%length;
        ListNode right = head;
        for(int i=0; i<k-1; i++){
            right = right.next;
        }
        ListNode left = new ListNode(-1);
        left.next = head;
        while(right.next!=null){
            left = left.next;
            right = right.next;
        }
        right.next = head; //重连
        ListNode res = left.next;
        left.next = null;
        return res;
    }
}