LeetCode刷题笔录 Rotate List

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:

Given

1->2->3->4->5->NULL

and k =

2

,

return

4->5->1->2->3->NULL

.

这题想法就是用两个pointer，一个比另外一个领先k个位置。对于上面那个例子，初始p1=1,p2=3；结束时p1=3,p2=5。

先扫描一遍链表，得到链表的长度，然后用k = n % length得到实际需要rotate的长度。

特殊情况：

1.list是空的

2.k=0

public class Solution {
public ListNode rotateRight(ListNode head, int n) {
if(head == null)
return null;
//get the length of the list
ListNode p1 = head;
int length = 1;
while(p1.next != null){
length++;
p1 = p1.next;
}
//get the rounded amount to rotate
int k = n % length;
//if k is zero, return as is
if(k == 0)
return head;
//use two pointers with k distance
p1 = head;
ListNode p2 = head;
for(int i = 0; i < k; i++){
p2 = p2.next;
}
while(p2.next != null){
p1 = p1.next;
p2 = p2.next;
}
//the new head pointer is p1.next
p2 = head;
ListNode newHead = p1.next;
p1.next = null;
ListNode p3 = newHead;
while(p3.next != null){
p3 = p3.next;
}
p3.next = p2;

return newHead;
}
}

可以优化一下：

第一遍跑到尾巴得到长度以后，直接p2.next = head，把尾巴和头连起来，然后再跑length - k长度，把next设置成null即可。这样少了一次扫描，也不用考虑k=0的情况了。