rotatelistjava_RotateList|Java最短代码实现

【思路】

1.    将尾指针指向头指针

2.    并将head指针向后移动k &＃61; count – k % count - 1位&＃xff0c;phead &＃61; head.next就是新链表的头指针

3.    将head.next置空&＃xff0c;并返回phead

public ListNode rotateRight(ListNode head, int k) {

if (head &＃61;&＃61; null) return head;

ListNode phead &＃61; head;

int count &＃61; 1;

while (phead.next !&＃61; null) {

count&＃43;&＃43;;

phead &＃61; phead.next;

}

k &＃61; count - k % count - 1;

phead.next &＃61; head;

while (k-- > 0)

head &＃61; head.next;

phead &＃61; head.next;

head.next &＃61; null;

return phead;

}

230 / 230test cases passed. Runtime:1 ms  Your runtime beats 14.33% of javasubmissions.

欢迎优化&＃xff01;