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].
Note: n
and m
both fit in range [0, 1000].
分情况来讨论:
- 当m和n其中有任意一个数是0时,返回1
- 当n = 1时
只有两种情况,0和1
- 当n = 2时,
这时候要看m的次数,如果m = 1,那么有三种状态 00,01,10
当m > 1时,那么有四种状态,00,01,10,11
- 当m = 1时,
此时n至少为3,那么我们有四种状态,000,010,101,011
- 当m = 2时,
此时n至少为3,我们有七种状态:111,101,010,100,000,001,110
- 当m > 2时,
此时n至少为3,我们有八种状态:111,101,010,100,000,001,110,011
class Solution {
public int flipLights(int n, int m) {
if (n == 0 || m == 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;
}
}