Rotate List

题目

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

2

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

方法

public ListNode rotateRight(ListNode head, int n) {
if (head == null) {
return null;
}
ListNode node = head;
int k = 0;
while (node != null) {
k++;
node = node.next;
}
int len = k;
n = n % len;
if (n == 0) {
return head;
}

ListNode start = new ListNode(0);
start.next = head;
ListNode pre = null;
ListNode end = null;
int i = 0;
end = start;
while (end.next != null) {

end = end.next;
if (pre != null) {
pre = pre.next;
}
i++;
if (i == n) {
pre = start;
}
}
if (i > n) {
end.next = start.next;
start.next = pre.next;
pre.next = null;
}
return start.next;
}