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:
cells.length == 8
cells[i]
is in{0, 1}
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;
}
}