Rotate List
Given a linked list, rotate the list to the right by k places, where k is non-negative.
Example 1:
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:
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
利用tail 和head来找到要rotate的部分链接的第一个 和最后一个,tail的下一个就是要旋转的第一个,head就是最后一个。因为从dummy开始走,head往前走k%size步以后,tail和head再一起走,head往前走的步数就是(size - k%size),这个步数走到的node正好是要旋转的链表开头的前一个node,也就是最后tail停在的位置,一开始想复杂了,被题目思维定式了,一个一个旋转,这样效率很低
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public int getLength(ListNode head){
if(head == null){
return 0;
}
int size = 0;
while(head!=null){
head = head.next;
size++;
}
return size;
}
public ListNode rotateRight(ListNode head, int k) {
if(head == null || k == 0){
return head;
}
int size = getLength(head);
k = k % size;
ListNode dummy = new ListNode(-1);
dummy.next = head;
head = dummy;
ListNode tail = dummy;
for(int i = 0; i < k; i++){
head = head.next;
}
while(head.next != null){
tail = tail.next;
head = head.next;
}
head.next = dummy.next;
dummy.next = tail.next;
tail.next = null;
return dummy.next;
}
}
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