1. 直接开新数组，暴力遍历，新位置=(原位置+k)%nums.length

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

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

2. 多轮元素轮转

还要想到 nums = {1,2,3,4}, k = 2的情况

class Solution {
    public void rotate(int[] nums, int k) {
        int start = 0, cnt = 0;
        while (cnt < nums.length) {
            int curPos = start;
            int curNum = nums[curPos];
            // 也可能 nums: 1 2 3 4, k = 2，直接跳回去的情况
            do {
                int nextPos = (curPos + k) % nums.length;
                int temp = nums[nextPos];
                nums[nextPos] = curNum;

                curNum = temp;
                curPos = nextPos;
                cnt++;
            }
            while (curPos != start); // curPos != satrt说明直接在这循环一次搞定
            start++;
        }
    }
}

3. 三次翻转

class Solution {
    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);
    }

    public void reverse(int[] nums, int start, int end) {
        int left = start, right = end;
        while (left < right) {
            int temp = nums[left];
            nums[left] = nums[right];
            nums[right] = temp;
            left++;
            right--;
        }
    }
}