[leetcode]957. Prison Cells After N Days 监狱房N天后的状态

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]

 

题意

给定一个长度固定为8的数组，表示一排监狱房的状态。 其中 1 表有人， 0 表无人， 根据规则求出N天后这排监狱房的状态

 

思路

1. 对于nextday的状态，很好判断。 思路类似[leetcode]605. Can Place Flowers能放花吗

2.N为7，还可以板着手指一天一天的推算结果。但是看到Example2的N为十亿，发现不是板着手指数这么简单。

3.试着将N的值扩大，看看会发生什么，试探将N扩大为20，发现，

N = 14 时候cells的状态跟N= 0 时候一模一样，

N = 15 时候cells的状态跟N= 1 时候一模一样，

N = 16 时候cells的状态跟N= 2 时候一模一样

...

也就是说，14的时候出现了一个循环节。

 

 

 

代码

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