_189.java

package com.fishercoder.solutions;

/**
 * 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?
 * */

public class _189 {

  public static class Solution1 {
    /**
     * O(n*k) time
     * O(1) space
     */
    public void rotate(int[] nums, int k) {
      int tmp;
      for (int i = 0; i < k; i++) {
        tmp = nums[nums.length - 1];
        for (int j = nums.length - 1; j > 0; j--) {
          nums[j] = nums[j - 1];
        }
        nums[0] = tmp;
      }
    }
  }

  public static class Solution2 {
    /**
     * using an extra array of the same size to copy it
     * O(n) time
     * O(n) space
     */
    public void rotate(int[] nums, int k) {
      int len = nums.length;
      int[] tmp = new int[len];
      for (int i = 0; i < len; i++) {
        tmp[(i + k) % len] = nums[i];
      }
      for (int i = 0; i < len; i++) {
        nums[i] = tmp[i];
      }
    }
  }

  public static class Solution3 {
    /**
     * reverse three times
     * O(n) time
     * O(1) space
     */
    public void rotate(int[] nums, int k) {
      int len = nums.length;
      k %= len;
      reverse(nums, 0, len - 1);
      reverse(nums, 0, k - 1);
      reverse(nums, k, len - 1);
    }

    private void reverse(int[] nums, int start, int end) {
      while (start < end) {
        int tmp = nums[start];
        nums[start] = nums[end];
        nums[end] = tmp;
        start++;
        end--;
      }
    }
  }
}