Rotate Array

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

Example 1:

Input: nums = [1,2,3,4,5,6,7], 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: nums = [-1,-100,3,99], 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]

Constraints:

• 1 <= nums.length <= 105

• -231 <= nums[i] <= 231 - 1

• 0 <= k <= 105

Follow up:

• Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.

• Could you do it in-place with O(1) extra space?

My Solutions:

方法1: 用另一个array储存结果，然后把结果复制到原来的array里。注意新的array是如何计算index的！

    public void rotate(int[] nums, int k) {
        int[] a = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            a[(i + k) % nums.length] = nums[i]; // 注意这里！把nums里的位置算好加入新的array
        }
        for (int j = 0; j < nums.length; j++) {
            nums[j] = a[j];
        }
    }

方法2: 把所有数字按照two pointers的方式reverse，找到需要切割的位置，两边再分别reverse一次，就能得到正确结果。

    public void rotate(int[] nums, int k) {
        if (nums == null || nums.length == 0 || k == 0) return;
        
        k = k % nums.length;
        reverse(nums, 0, nums.length - 1);
        reverse(nums, 0, k - 1); // 注意这里！因为reverse的时候是【】包含的
        reverse(nums, k, nums.length - 1);
    }
    
    private void reverse(int[] nums, int start, int end) {
        while (start < end) {
            int temp = nums[start];
            nums[start] = nums[end];
            nums[end] = temp;
            start++;
            end--;
        }
    }