Rotate Array Leetcode

Rotate an array of n elements to the right by k steps.

For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4]. 

Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.

这道题要求最好用O(1)做出来，我一开始是新建了一个数组，这个reverse三次的方法也不知道怎么找出来的，挺神奇的。

public class Solution {
    public void rotate(int[] nums, int k) {
        if (nums == null || nums.length == k || nums.length <= 1) {
            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 left, int right) {
        while (left < right) {
            int tmp = nums[right];
            nums[right] = nums[left];
            nums[left] = tmp;
            left++;
            right--;
        }
    }
}

 

重温这道题，当时没有理解好像，再做还是新建数组。。。

public class Solution {
    public void rotate(int[] nums, int k) {
        if (nums == null) {
            return;
        }
        int n = nums.length;
        k = k % n;
        int[] temp = new int[k];
        int i;
        for (i = nums.length - 1; i >= nums.length - k; i--) {
            temp[i + k - nums.length] = nums[i];
        }
        for (int j = i; j >= 0; j--) {
            nums[j + k] = nums[j];
        }
        for (int m = k - 1; m >= 0; m--) {
            nums[m] = temp[m];
        }
    }
}

又看了一下，好像是直接翻转就可以了。主要是找规律。代码和之前的基本一样，就不放了。