Leetcode 189. Rotate Array

Leetcode 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.

[show hint]

Hint:
Could you do it in-place with O(1) extra space?
Related problem: Reverse Words in a String II

1.use extra space

public void rotate(int[] nums, int k) {
        if (nums.length == 0)
            return;
        int m = k % nums.length;
        int j = 0;
        int[] newNum = new int[nums.length];
        for(int i = nums.length-m; i < nums.length; i++){
            newNum[j] = nums[i];
            j++;
        }
        for(int i = 0; i < nums.length-m; i++){
            newNum[j] = nums[i];
            j++;
        }
        for(int i = 0; i < nums.length; i++)
            nums[i] = newNum[i];
    }

2.翻转字符：

public void rotate(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--;
    }
}

3.映射关系 :

public void Rotate(int[] nums, int k) {
    if(nums.Length == 0 || k % nums.Length == 0) return;
    int start = 0, i = start, curNum = nums[i], count = 0;
    while(count < nums.Length){
        i = (i + k) % nums.Length;
        int tmp = nums[i];
        nums[i] = curNum;
        if(i == start){
            start++;
            i = start;
            curNum = nums[i];
        }
        else curNum = tmp;
        count++;
    }
}

4.通过不停的交换某两个数字的位置来实现旋转

public void rotate(int[] nums, int k) {
   if (nums.length == 0) return;
    int n = nums.length;
    while ((k %= n) > 0 && n > 1) {
        int range = n - k;
        for (int i = 1; i <= range; i++) {
            int val = nums[n - i];
            nums[n - i] = nums[n - i - k];
            nums[n - i - k] = val;
        }
        n = k;
        k = n - (range % k);
    }
}

1 2 3 4 5 6 7
5 2 3 4 1 6 7
5 6 3 4 1 2 7
5 6 7 4 1 2 3
5 6 7 1 4 2 3
5 6 7 1 2 4 3
5 6 7 1 2 3 4