[LC] 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]

Brute Force -> LTE
class Solution {
    public int[] prisonAfterNDays(int[] cells, int N) {
        int[][] arr = new int[N + 1][cells.length];
        for (int i = 0; i < cells.length; i++) {
            arr[0][i] = cells[i];
        }
        for (int i = 1; i <= N; i++) {
            for (int j = 0; j < cells.length; j++) {
                if (j == 0 || j == cells.length - 1) {
                    continue;
                }
                if (arr[i - 1][j - 1] == 0 && arr[i - 1][j + 1] == 0 || arr[i - 1][j - 1] == 1 && arr[i - 1][j + 1] == 1) {
                    arr[i][j] = 1;
                }
            }
            // System.out.println(i + ": " + Arrays.toString(arr[i]));
        }
        return arr[N];
    }
}

solution: find there is cycle form

class Solution {
    public int[] prisonAfterNDays(int[] cells, int N) {
        Set<String> set = new HashSet<>();
        boolean hasCycle = false;
        int count = 0;
        int[] nxt = new int[cells.length];
        for (int i = 0; i < N; i++) {
            nxt = getNext(cells);
            String str = Arrays.toString(nxt);
            if (set.contains(str)) {
                hasCycle = true;
                break;
            } else {
                set.add(str);
                count += 1;
            }
            cells = nxt;
        }
        if (hasCycle) {
            N = N % count;
            for (int i = 0; i < N; i++) {
                cells = getNext(cells);
            }
        }
        return cells;
    }
    
    private int[] getNext(int[] cells) {
        int[] arr = new int[cells.length];
        for (int i = 1; i < cells.length - 1; i++) {
           arr[i] = cells[i - 1] == cells[i + 1] ? 1 : 0; 
        }
        return arr;
    }
}
原文地址:https://www.cnblogs.com/xuanlu/p/12702783.html