957. 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]

class Solution {
    public int[] prisonAfterNDays(int[] cells, int N) {
        Set<String> set = new HashSet();
        int cycle = 0;
        boolean hascyc = false;
        for(int i = 1; i <= N; i++){
            int[] next = help(cells);
            String s = Arrays.toString(next);
            if(set.contains(s)){
                hascyc = true;
                break;
            }
            else{
                cycle++;
                set.add(s);
            }
            cells = next;   
        }
        if(!hascyc) return cells;
        else{
            N %= cycle;
            for(int i = 0; i < N; i++){
                cells = help(cells);
            }
            return cells;
        }
    }
    public int[] help(int[] cells){
        int[] next = new int[cells.length];
        for(int i = 1; i < cells.length - 1; i++){
            next[i] = cells[i-1] == cells[i+1] ? 1: 0;
        }
        return next;
    }
}

simulating，

先读题，一个cell左边右边是相同状态下一次这个cell就是1，否则是0

因为只有8个cell，充其量就2^8 == 256种，超过了肯定有循环

用hashset判断有无循环，有就把N mod变小，重新call help方法即可