package leetcode_1To300;

/**
 * 本代码来自 Cspiration，由 @Cspiration 提供
 * 题目来源：http://leetcode.com
 * - Cspiration 致力于在 CS 领域内帮助中国人找到工作，让更多海外国人受益
 * - 现有课程：Leetcode Java 版本视频讲解（1-900题）（上）（中）（下）三部
 * - 算法基础知识（上）（下）两部；题型技巧讲解（上）（下）两部
 * - 节省刷题时间，效率提高2-3倍，初学者轻松一天10题，入门者轻松一天20题
 * - 讲师：Edward Shi
 * - 官方网站：https://cspiration.com
 * - 版权所有，转发请注明出处
 */
public class _189_RotateArray {
    /**
     * 189. Rotate Array
     * 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.

     * @param nums
     * @param k
     */
    //time : O(n)  space : O(n)
    public void rotate(int[] nums, int k) {
        int[] temp = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            temp[(i + k) % nums.length] = nums[i];
        }
        for (int i = 0; i < nums.length; i++) {
            nums[i] = temp[i];
        }
    }

    //time : O(n)  space : O(1)

    /**
     Original List                   : 1 2 3 4 5 6 7
     After reversing all numbers     : 7 6 5 4 3 2 1
     After reversing first k numbers : 5 6 7 4 3 2 1
     After revering last n-k numbers : 5 6 7 1 2 3 4 --> Result
     * @param nums
     * @param k
     */
    public void rotate2(int[] nums, int 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) {
        while (start < end) {
            int temp = nums[start];
            nums[start] = nums[end];
            nums[end] = temp;
            start++;
            end--;
        }
    }
}