05 August 2008
Given a linked list, rotate the list to the right by k places, where k is non-negative.
Example 1:
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Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL
Example 2:
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Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL
一个链表把右边k位的结点挪到左边来，可以分三步来做：1）计算链表长度；2）定位到len - k % len的位置因为k可能比len大，所有是k % len；3）在定位好的位置处开始rotate。
时间O(n)，空间O(1)。
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/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode rotateRight(ListNode head, int k) {
if (head == null || head.next == null) {
return head;
}
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode fast = dummy;
ListNode slow = dummy;
//找到链表的长度
int len;
for (len = 0; fast.next != null; len++) {
fast = fast.next;
}
//定位到len - k%len的位置
for (int i = 0; i < len - k%len; i++) {
slow = slow.next;
}
//开始做rotation
fast.next = dummy.next;
dummy.next = slow.next;//把i-k%i的节点放到dummy之后（return后的第一位）
slow.next = null;
return dummy.next;
}
}
也可以把链表弄成一个环来做
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/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode rotateRight(ListNode head, int k) {
if (head == null || head.next == null) {
return head;
}
ListNode index = head;
int len = 1;
while (index.next != null) {
index = index.next;
len++;
}
index.next = head;//形成环
for (int i = 1; i < len - k % len; i++) {
head = head.next;
}
ListNode result = head.next;
head.next = null;//再断开环形成新的单链表
return result;
}
}