0957. Prison Cells After N Days (M)

题目

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

题意

给定一个一维数组cells，进行N次循环，每次循环中如果cells[i]邻接的两个元素都为1或都为0，则cells[i]变为1，否则cells[i]变为0。首元素和末元素只会变为0。求N次循环后数组的状态。

思路

和 0289. Game of Life (M) 有点类似，每次状态的变更都是同时发生的，不能用前一个变化后的值去影响后一个将要改变的值。每次循环可以直接复制一个数组进行操作。同时，打表可以发现数组的状态是有规律的，以14作为一个周期，因此N等效于(N-1)%14+1。

代码实现

Java

class Solution {
    public int[] prisonAfterNDays(int[] cells, int N) {
        N = (N - 1) % 14 + 1;
        for (int i = 1; i <= N; i++) {
            int[] aux = Arrays.copyOf(cells, 8);
            for (int j = 1; j < 7; j++) {
                cells[j] = aux[j - 1] == aux[j + 1] ? 1 : 0;
            }
            cells[0] = cells[7] = 0;
        }
        return cells;
    }
}