[LeetCode] 189. Rotate Array

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Problem

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

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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:

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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?

Explanation 1

  1. We can create a copy of the array temp, then iterate the array with a relationship (i+k)%n -> i.

Solution 1

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class Solution {
    public void rotate(int[] nums, int k) {
        int[] temp = new int[nums.length];
        System.arraycopy(nums, 0, temp, 0, nums.length);

        for (int i = 0; i < nums.length; i++) {
            nums[(i+k) % nums.length] = temp[i];
        }
    }
}

Explanation 2

  1. We can use $O(1)$ space to solve this problem.

  2. First, reverse the first (n-k) numbers.

  3. Then, reverse the last k numbers.

  4. Finally, reverse the whole array.

Solution 2

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class Solution {
    void reverse(int[] nums, int left, int right) {
        while (left < right) {
            int temp = nums[left];
            nums[left] = nums[right];
            nums[right] = temp;
            left += 1;
            right -= 1;
        }
    }

    public void rotate(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            throw new IllegalArgumentException("Illegal Argument");
        }
        k = k % nums.length;
        reverse(nums, 0, nums.length-k-1);
        reverse(nums, nums.length-k, nums.length-1);
        reverse(nums, 0, nums.length-1);
    }
}