题目描述
示例1
输入：
n = 1, m = 1.
输出：
2
解释：
状态为: [开], [关]
示例2
输入：
n = 2, m = 1.
输出：
3
解释：
状态为: [开, 关], [关, 开], [关, 关]
示例3
输入：
n = 3, m = 1.
输出：
4
解释：
状态为: [关, 开, 关], [开, 关, 开], [关, 关, 关], [关, 开, 开].
提示
题解
代码
c++
class Solution {
public:
int flipLights(int n, int m) {
if (m == 0) return 1;
if (n == 1) return 2;
m = min(m, 3);
if (n == 2) return vector<int>{3, 4, 4}[m-1];
return vector<int>{4, 7, 8}[m-1];
}
};
python
class Solution:
def flipLights(self, n: int, m: int) -> int:
if m == 0: return 1
if n == 1: return 2
m = min(m, 3)
if n == 2: return [3, 4, 4][m-1]
return [4, 7, 8][m-1]
python（枚举）
class Solution:
def flipLights(self, n, m):
seen = set()
for cand in itertools.product((0, 1), repeat = 4):
if sum(cand) % 2 == m % 2 and sum(cand) <= m:
A = []
for i in range(min(n, 3)):
light = 1
light ^= cand[0]
light ^= cand[1] and i % 2
light ^= cand[2] and i % 2 == 0
light ^= cand[3] and i % 3 == 0
A.append(light)
seen.add(tuple(A))
return len(seen)
