Rotate Array（189 秘通，61）
Rotate an array of n elements to the right by k steps.
For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].
Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
Related problem: Reverse Words in a String II
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.
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三步：全部反轉(zhuǎn)为严；反轉(zhuǎn)序列1 ；反轉(zhuǎn)序列2
public class Solution {
public void rotate(int[] nums, int k) {
k %= nums.length;
reverse(nums, 0, nums.length - 1);
reverse(nums, 0, k - 1);
reverse(nums, k, nums.length - 1);
}
public void reverse(int[] nums, int start, int end) {
while (start < end) {
int temp = nums[start];
nums[start] = nums[end];
nums[end] = temp;
start++;
end--;
}
}
}
Rotate List（61）
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.
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Since n may be a large number compared to the length of list. So we need to know the length of linked list.After that, move the list after the (l-n%l )th node to the front to finish the rotation.
Ex: {1,2,3} k=2 Move the list after the 1st node to the front
Ex: {1,2,3} k=5, In this case Move the list after (3-5%3=1)st node to the front.
So the code has three parts.
- Get the length
- Move to the (l-n%l)th node
3)Do the rotation
public ListNode rotateRight(ListNode head, int n) {
if (head==null||head.next==null) return head;
ListNode dummy=new ListNode(0);
dummy.next=head;
ListNode fast=dummy,slow=dummy;
int i;
for (i=0;fast.next!=null;i++)//Get the total length
fast=fast.next;
for (int j=i-n%i;j>0;j--) //Get the i-n%i th node
slow=slow.next;
fast.next=dummy.next; //Do the rotation
dummy.next=slow.next;
slow.next=null;
return dummy.next;
}
