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

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三段翻转。
Time: O(n)
Space: O(n)

public class Solution {
    public void rotate(int[] nums, int k) {
        if (nums.length == 0 || k % nums.length == 0) return;
        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 l, int r) {
        while (l < r) {
            int temp = nums[l];
            nums[l++] = nums[r];
            nums[r--] = temp;
        }
        return;
    }
}

正常做得用个O(n)的数据结构。

GOOGLE有个SHIFT的题，follow up要求in-place without 三段。。

试一试。。

1 2 3 4 5 6 7 = > 5 6 7 1 2 3 4
1 2 3 1 5 6 7 t = 4
1 2 3 1 5 6 4 t = 7
1 2 7 1 5 6 4 t = 3
1 2 7 1 5 3 4 t = 6
1 6 7 1 5 3 4 t = 2
1 6 7 1 2 3 4 t = 5
5 6 7 1 2 3 4..done..
从第一个开始，每次移K个位置，总共移动nums.length个

这个CASE表明移nums.length个不行的。。会回到原点
1 2 3 4 = > 3 4 1 2
1 2 1 4 t = 3

所以回到远点我们就到下一个。。总共还是nums.length

嘿嘿。。做出来了

public class Solution {
    public void rotate(int[] nums, int k) {
        if (nums.length == 0 || k % nums.length == 0) return;
        k %= nums.length;
        int done = 0;
        for (int i = 0; i < k; i++) {
            int j = i;
            int tempVal = nums[i];
            while (done < nums.length) {
                int tempJ = (j + k) % nums.length;
                int temp = nums[tempJ];
                nums[tempJ] = tempVal;
                tempVal = temp;
                j = tempJ;
                done++;
                if (j == i) break;
            }
        }
    }
}