灯的开关 Bulb Switcher II

问题：

There is a room with n lights which are turned on initially and 4 buttons on the wall. After performing exactly m unknown operations towards buttons, you need to return how many different kinds of status of the n lights could be.

Suppose n lights are labeled as number [1, 2, 3 ..., n], function of these 4 buttons are given below:

1. Flip all the lights.
2. Flip lights with even numbers.
3. Flip lights with odd numbers.
4. Flip lights with (3k + 1) numbers, k = 0, 1, 2, ...

Example 1:

Input: n = 1, m = 1.
Output: 2
Explanation: Status can be: [on], [off]

Example 2:

Input: n = 2, m = 1.
Output: 3
Explanation: Status can be: [on, off], [off, on], [off, off]

Example 3:

Input: n = 3, m = 1.
Output: 4
Explanation: Status can be: [off, on, off], [on, off, on], [off, off, off], [off, on, on].

Note: n and m both fit in range [0, 1000].

解决：

① 仍是找规律。

咱们只须要考虑当 n<=2 and m < 3 的特殊情形。由于当 n >2 and m >=3, 结果确定是 8.

四个按钮的功能：

• 翻转全部的灯。
• 翻转偶数的灯。
• 翻转奇数的灯。
• 翻转（3k + 1）数字，k = 0，1，2，...

若是咱们使用按钮1和2，则等同于使用按钮3。

一样的：

1 + 2 → 3,1 + 3 → 2，2 + 3 → 1
因此，只有8种结果：1，2，3，4，1 + 4，2 + 4，3 + 4，当n> 2和m> = 3时，咱们能够获得全部的状况。

class Solution { //7ms
    public int flipLights(int n, int m) {
        if (m == 0) return 1;
        if (n == 1) return 2;
        if (n == 2 && m == 1) return 3;
        if (n == 2) return 4;
        if (m == 1) return 4;
        if (m == 2) return 7;
        if (m >= 3) return 8;
        return 8;
    }
}

② 

//O(1)数学问题，总共有8个state  1111,1010,0101,0111,0000,0011, 1100 and 1001.
//须要枚举 n>3之后就只能是这8个state了
//n == 1 Only 2 possibilities: 1 and 0.
//n == 2 After one operation, it has only 3 possibilities: 00, 10 and 01. After two and more operations, it has only 4 possibilities: 11, 10, 01 and 00.
//n == 3 After one operation, it has only 4 possibilities: 000, 101, 010 and 011. After two operations, it has 7 possibilities: 111,101,010,100,000,001 and 110. After three and more operations, it has 8 possibilities, plus 011 on above case.
//n >= 4 After one operation, it has only 4 possibilities: 0000, 1010, 0101 and 0110.
//After two or more operations: it has 8 possibilities, 1111,1010,0101,0111,0000,0011, 1100 and 1001.
class Solution {//8ms
    public int flipLights(int n, int m) {
        n = Math.min(n, 3);
        if (m == 0) return 1;
        if (m == 1) return n == 1 ? 2 : n == 2 ? 3 : 4;
        if (m == 2) return n == 1 ? 2 : n == 2 ? 4 : 7;
        return n == 1 ? 2 : n == 2 ? 4 : 8;
    }
}