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Prison Cells After N Days-LeetCode#957

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]

Note:

  1. cells.length == 8
  2. cells[i] is in {0, 1}
  3. 1 <= N <= 10^9
思路:连续8个牢房,每个牢房被占用或空置。每天,根据以下规则,单元是否被占用或空置变化:如果一个小区有两个相邻的邻居,它们都被占用或者都是空的,那么该小区就会被占用。否则,它会变空。请注意,因为监狱是一排,所以行中的第一个和最后一个单元格不能有两个相邻的邻居。我们用以下方式描述监狱的当前状态:如果第i个cell被占用,则cell [i] == 1,否则cells [i] == 0。鉴于监狱的初始状态,在N天之后返回监狱的状态(以及上述N个这样的改变。)
看到N=1000000000,思考一下应该是有循环的
  • 0 1 0 1 1 0 0 1
  • 0 1 1 0 0 0 0 0
  • 0 0 0 0 1 1 1 0
  • 0 1 1 0 0 1 0 0
  • 0 0 0 0 0 1 0 0
  • 0 1 1 1 0 1 0 0
  • 0 0 1 0 1 1 0 0
  • 0 0 1 1 0 0 0 0
  • 0 0 0 0 0 1 1 0
  • 0 1 1 1 0 0 0 0
  • 0 0 1 0 0 1 1 0
  • 0 0 1 0 0 0 0 0
  • 0 0 1 0 1 1 1 0
  • 0 0 1 1 0 1 0 0
  • 0 0 0 0 1 1 0 0
  • 0 1 1 0 0 0 0 0

可以看出周期为14。

代码如下:

public class PrisonAfterNDays {
    public static int[] prisonAfterNDays(int[] cells, int N) {
        int n = 0;
        List<int[]> list = new ArrayList<>();
        switchCells(cells, list, n);
        if (N < 14){
            return list.get(N - 1);
        }else {
            if (N % 14 == 0){
                return list.get(13);
            }else {
                return list.get(N % 14 - 1);
            }
        }

    }

    static void switchCells(int[] cells, List<int[]> list, int n){
        n++;
        int[] temp = new int[8];
        temp[0] = 0;
        temp[7] = 0;
        for (int i = 1; i < cells.length - 1; i++) {
            temp[i] = cells[i - 1] == cells[i + 1] ? 1 : 0;
        }
        list.add(temp);
        if (n < 14){
            switchCells(temp, list, n);
        }
    }
}

 

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