Question
Formatted question description: https://leetcode.ca/all/189.html
189. Rotate Array
Given an array, rotate the array to the right by k steps, where k is non-negative.
Example 1:
Input: [1,2,3,4,5,6,7] and k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]
Example 2:
Input: [-1,-100,3,99] and k = 2
Output: [3,99,-1,-100]
Explanation:
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]
Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
Could you do it in-place with O(1) extra space?
@tag-array
Algorithm
O(1) extra space.
Similar to the method of flipping characters, the idea is to flip the first n-k numbers first, then flip the last k numbers, and finally flip the entire array.
[1,2,3,4,5,6,7] => [4,3,2,1,5,6,7]
[4,3,2,1,5,6,7] => [4,3,2,1,7,6,5]
[4,3,2,1,7,6,5] => [5,6,7,1,2,3,4]
Code
Java
import org.apache.commons.lang3.ArrayUtils;
public class Rotate_Array {
class Solution {
public void rotate(int[] nums, int k) {
if (nums.length == 0 || (k %= nums.length) == 0) {
return;
}
int n = nums.length;
// [1,2,3,4,5,6,7] => [4,3,2,1,5,6,7]
ArrayUtils.reverse(nums, 0, n - k); // @note: org.apache.commons.lang3.ArrayUtils
// [4,3,2,1,5,6,7] => [4,3,2,1,7,6,5]
ArrayUtils.reverse(nums, n - k, nums.length);
// [4,3,2,1,7,6,5] => [5,6,7,1,2,3,4]
ArrayUtils.reverse(nums, 0, nums.length);
}
}
class Solution_kTimesAllItemsMoving {
public void rotate(int[] nums, int k) {
while ( k > 0) {
k--;
int tmp = nums[nums.length - 1]; // last one
for (int i = nums.length - 1; i >= 1; i--) {
nums[i] = nums[i - 1];
}
nums[0] = tmp;
}
}
}
}