leetcode 之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.

[show hint]

Related problem: Reverse Words in a String II

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

好的~，这题在编程珠玑还是编程之美里看过。不记得了。反正题意记着。

诀窍在于两次翻转

int len = nums.length;        if (len == 1 || k == 0 || len == k) {            return;        }        k = k % len;        for (int i = 0; i < (len - k) / 2; i++) {            int t = nums[i];            nums[i] = nums[len - k - i - 1];            nums[len - k - i - 1] = t;        }        for (int i = len - k, j = 1; i < len - k/ 2; i++) {            int t = nums[i];            nums[i] = nums[len - j];            nums[len - j] = t;            j++;        }        for (int i = 0; i < len / 2; i++) {            int t = nums[i];            nums[i] = nums[len - i - 1];            nums[len - i - 1] = t;        }

它说有三种方法。这是第一种，其它想到再说。
第二种

   public void rotate(int[] nums, int k) {        int len = nums.length;        k = k % len;        if (len == 1 || k == 0 || len == k) {            return;        }        int[] result=new int[len];        for (int i = 0; i <len; i++) {            if (i<k) {                result[k-i-1]=nums[len-i-1];            }else{                result[i]=nums[i-k];            }        }        for(int i=0;i<len;i++){            nums[i]=result[i];        }    }