189 Rotate Array

public class Solution {
    public void rotate(int[] nums, int k) {
        int n = nums.length;
        k = k % n;
        int[] res = new int[n];
        
        for (int i = 0; i < n; i++) {
            res[(i + k)%n] = nums[i];
        }
        for (int i = 0; i < n; i++) {
            nums[i] = res[i];
        }
    }
}

这样space complexsion 不是O(1)

时间空间复杂度：http://blog.csdn.net/zolalad/article/details/11848739

参考答案：

直接观察规律，类似三部反转法。

    public void rotate(int[]  nums, int k) {
        if (nums.length == 0) {
            return;
        }
        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);
        
    }
    
    public void reverse(int[] nums, int start, int end) {
        while (start < end) {
            int tmp = nums[start];
            nums[start] = nums[end];
            nums[end] = tmp;
            start++;
            end--;
        }
    }