There are 8 prison cells in a row, and each cell is either occupied or vacant.
Each day, whether the cell is occupied or vacant changes according to the following rules:
- If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied.
- Otherwise, it becomes vacant.
(Note that because the prison is a row, the first and the last cells in the row can't have two adjacent neighbors.)
We describe the current state of the prison in the following way: cells[i] == 1 if the i-th cell is occupied, else cells[i] == 0.
Given the initial state of the prison, return the state of the prison after N days (and N such changes described above.)
Example 1:
Input: cells = [0,1,0,1,1,0,0,1], N = 7
Output: [0,0,1,1,0,0,0,0]
Explanation:
The following table summarizes the state of the prison on each day:
Day 0: [0, 1, 0, 1, 1, 0, 0, 1]
Day 1: [0, 1, 1, 0, 0, 0, 0, 0]
Day 2: [0, 0, 0, 0, 1, 1, 1, 0]
Day 3: [0, 1, 1, 0, 0, 1, 0, 0]
Day 4: [0, 0, 0, 0, 0, 1, 0, 0]
Day 5: [0, 1, 1, 1, 0, 1, 0, 0]
Day 6: [0, 0, 1, 0, 1, 1, 0, 0]
Day 7: [0, 0, 1, 1, 0, 0, 0, 0]
Example 2:
Input: cells = [1,0,0,1,0,0,1,0], N = 1000000000
Output: [0,0,1,1,1,1,1,0]
Note:
cells.length == 8cells[i]is in{0, 1}1 <= N <= 10^9
如果知道有规律，那么就可以好解，只需要暴力跑一次找出规律就行。这题还真的是有规律，就是14天一次的规律
class Solution {
public:
vector<int> prisonAfterNDays(vector<int>& cells, int N) {
int len = cells.size();
int num[];
int x = N% == ? : N%;
for(int i=;i<x;i++){
for(int j=;j<=;j++){
num[j] = cells[j-]+cells[j+];
}
for(int j=;j<=;j++){
if(num[j]==||num[j]==){
cells[j]=;
}else{
cells[j]=;
}
}
if(cells[]==){
cells[]=;
}
if(cells[]==){
cells[]=;
}
}
//89
//96
//14
//
return cells;
}
};