LeetCode - 189. Rotate Array

题目：

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.

思路与步骤：

思路1：

用一个 List 存储变换好顺序的数组，然后赋值给原数组；

思路2：

用一个长度为 k 的数组存储变换好的数组的前半截，后半截直接在原数组上修改；

思路3：

写一个 reverse() 函数，然后用三步逆转实现。

编程实现：

Solution-1：

public class Solution {
    public void rotate(int[] nums, int k) {
        int len = nums.length;
        if(len <= 1) return;
        k = k % len;
        List<Integer> numlist = new ArrayList<Integer>();
        for(int i=0; i<k; i++)    numlist.add(nums[nums.length-k+i]);
        for(int i=k; i<nums.length; i++)  numlist.add(nums[i-k]);
        for(int i=0; i<nums.length; i++)    nums[i] = numlist.get(i);
    }
}

Solution-2：

public class Solution {
    public void rotate(int[] nums, int k) {
        if(nums.length <= 1) return;
        int step = k % nums.length;
        int[] tmp = new int[step];
        for(int i=0; i<step; i++)   tmp[i] = nums[nums.length - step + i];
        for(int i=nums.length-step-1; i>= 0; i--)    nums[i + step] = nums[i];
        for(int i=0; i<step; i++)   nums[i] = tmp[i];
    }
}

Solution-3：

public class Solution {
    public void rotate(int[] nums, int k) {
        int len = nums.length;
        if(len <= 1) return;
        k = k % len;
        reverse(nums, 0, len-1);
        reverse(nums, 0, k-1);
        reverse(nums, k, len-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--;
        }
    }
}