Rotate Array

项目github地址：bitcarmanlee easy-algorithm-interview-and-practice
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Problem

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]. How many different ways do you know to solve this problem?

这道题是旋转数组。大致思路如下：
1.复制数组，最简单的方式。
2.采用类似冒泡排序的方式，每次将最右边一个移到最前面。
3.采用旋转逆序的方式。

talk is cheap,show me the code!

public class RotateArray {public static void printArray(int[] array) {for(int i=0; i<array.length; i++) {System.out.print(array[i] + " ");}}//方法一:复制数组,时间复杂度与空间复杂度均为O(n)public static void rotate(int[] nums, int k) {if(k > nums.length) {k%= nums.length;}int[] result = new int[nums.length];for(int i=0; i<k; i++) {result[i] = nums[nums.length-k+i];}int j = 0;for(int i=k; i<nums.length; i++) {result[i] = nums[j];j++;}System.arraycopy(result,0,nums,0,nums.length);}//方法二:类似于冒泡排序,时间复杂度O(nk),空间复杂度O(1)public static void bubble_rotate(int[] nums, int k) {if(nums == null || k < 0) {throw new IllegalArgumentException("Illegal argument!");}for(int i=0; i<k; i++) {for(int j=nums.length-1; j>0; j--) {int tmp = nums[j];nums[j] = nums[j-1];nums[j-1] = tmp;}}}//方法三:逆序的方式,时间复杂度O(n),空间复杂度O(1)public static void reverse_rotate(int[] nums, int k) {if(nums == null || k < 0) {throw new IllegalArgumentException("Illegal argument!");}reverse_array(nums,0,k-1);reverse_array(nums,k,nums.length-1);reverse_array(nums,0,nums.length-1);}public static void reverse_array(int[] array, int left, int right) {if (array == null || array.length == 0 || left < 0 || right < 0) {throw new IllegalArgumentException("Illegal argument!");}while(left < right) {int tmp = array[left];array[left] = array[right];array[right] = tmp;left++;right--;}}public static void main(String[] args) {int[] nums = {1,2,3,4,5,6,7};reverse_rotate(nums,3);printArray(nums);}
}

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