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?

class Solution {
    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];
        }
        for (int i = 0; i < nums.length; i++) {
            nums[i] = a[i];
        }
    }
}

鬼才，用i+k余length解决先copy后k个数据

public class Solution {
    public void rotate(int[] nums, int k) {
        k %= nums.length;
        reverse(nums, 0, nums.length - k);
        reverse(nums, nums.length - k, nums.length);
        reverse(nums, 0, nums.length);
    }
    private static void reverse(int[] nums, int begin, int end) {
        int left = begin;
        int right = end - 1;
        while (left < right) {
            // swap
            int tmp = nums[left];
            nums[left] = nums[right];
            nums[right] = tmp;
            ++left;
            --right;
        }
    }
}

方法2，reverse3次，分别是先将数组分为两段，前n-k个为一段，后k个元素作为第二段，将第一段reverse, 第二段 reverse, 然后将整个数组reverse, 这样经过三轮reverse，就完成了循环右移。时间复杂度O(n)，空间复杂度O(1)。

注意有k % = nums.length防止k 大于 length