189. Rotate Array -- 将数组前一半移到后一半

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.

（1）

/*

 * this solution is so-called three times rotate method

 * because (X^TY^T)^T = YX, so we can perform rotate operation three times to get the result

 * obviously, the algorithm consumes O(1) space and O(n) time

 */

void rotate(int nums[], int n, int k) {

    if (k<=) return;

    k %= n; 

    reverseArray(nums, n-k, n-);

    reverseArray(nums, , n-k-);

    reverseArray(nums, , n-);    

}

（2）

/*

 * How to change [0,1,2,3,4,5,6] to [4,5,6,0,1,2,3] by k = 3?

 *

 * We can change by following rules: 

 *

 *     [0]->[3], [3]->[6], [6]->[2],  [2]->[5], [5]->[1], [1]->[4]

 *    

 *

 */

void rotate(int nums[], int n, int k) {

    if (k<=) return;

    k %= n;

    int currIdx=, newIdx=k;

    int tmp1 = nums[currIdx], tmp2; 

    int origin = ;

    for(int i=; i<n; i++){

        tmp2 = nums[newIdx];

        nums[newIdx] = tmp1;

        tmp1 = tmp2; 

        currIdx = newIdx;

        //if we meet a circle, move the next one

        if (origin == currIdx) {

            origin = ++currIdx;

            tmp1 = nums[currIdx];

        }

        newIdx = (currIdx + k) % n;

    } 

}