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

Example：

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]

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?

解釋下題目：

數(shù)組的數(shù)字往后推移k位，如果超出了最后則接到頭部

1. 倒置的方法

實際耗時：0ms

    public void rotate(int[] nums, int k) {
        k = k % nums.length;
        reverse(nums, 0, nums.length - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, nums.length - 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--;
        }
    }

踩過的坑：[-1] k=2

??我自己想到的思路是有三種，第一種是，搞個函數(shù)，然后這個函數(shù)的作用是每次給數(shù)組后退1格，然后如果是k的話就重復(fù)k次，這樣時間是O(nk) 不需要額外的空間。第二種是再開辟一個同樣大小的數(shù)組，很簡單。最后一種就是上面代碼里的，其實這種rotate的題目最后都可以用倒序來實現(xiàn)。

時間復(fù)雜度O(n)

空間復(fù)雜度O(1)