[Leetcode] 672. Bulb Switcher II 解题报告

题目：

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:
Flip all the lights.
Flip lights with even numbers.
Flip lights with odd numbers.
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].

思路：

n小于等于1的时候，属于平凡情况，可以直接返回。对于这四种类型的操作，我们可以得出如下的观察结论：

1）对于每种类型的操作，每操作两次就等于没有操作，所以其操作结果只与其操作次数的奇偶性有关。

2）前三种操作可以被规约为1次或者0次操作。例如，操作1和操作2的结果就等于操作3，而操作1，操作2和操作3的结果就是保持原貌。

3）对于n > 3的情况，其结果与n == 3的情况相同，因为当前3个灯泡的状态确定之后，后面的所有的灯泡的状态都是确定的。

所以，我们可以将所有的情况规约为m <= 3, n <= 3的情况。所以就可以直接返回结果了，时间复杂度和空间复杂度都是O(1)。

代码：

class Solution {
public:
int flipLights(int n, int m) {
if (m == 0 || n == 0) {
return 1;
}
if (n == 1) {
return 2;
}
if (n == 2) {
return m == 1 ? 3 : 4;
}
if (m == 1) {
return 4;
}
return m == 2 ? 7 : 8;
}
};