115th LeetCode Weekly Contest Prison Cells After N Days

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:

1. cells.length == 8
2. cells[i] is in {0, 1}
3. 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[10];
        int x = N%14 == 0 ? 14: N%14;
        for(int i=0;i<x;i++){
            
            
            for(int j=1;j<=6;j++){
                num[j] = cells[j-1]+cells[j+1];
            }
            for(int j=1;j<=6;j++){
                if(num[j]==2||num[j]==0){
                    cells[j]=1;
                }else{
                    cells[j]=0;
                }
            }
            if(cells[0]==1){
                cells[0]=0;
            }
            if(cells[7]==1){
                cells[7]=0;
            }
        }
        //89
        //96
        //14
        //199
        return cells;
    }
};