(Java) LeetCode 289. Game of Life —— 生命游戏

According to the Wikipedia's article: "The Game of Life, also known simply as Life, is a cellular automaton devised by the British mathematician John Horton Conway in 1970."

Given a board with m by n cells, each cell has an initial state live (1) or dead (0). Each cell interacts with its eight neighbors (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article):

1. Any live cell with fewer than two live neighbors dies, as if caused by under-population.
2. Any live cell with two or three live neighbors lives on to the next generation.
3. Any live cell with more than three live neighbors dies, as if by over-population..
4. Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.

Write a function to compute the next state (after one update) of the board given its current state. The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously.

Example:

Input: 
[
  [0,1,0],
  [0,0,1],
  [1,1,1],
  [0,0,0]
]
Output: 
[
  [0,0,0],
  [1,0,1],
  [0,1,1],
  [0,1,0]
]

Follow up:

1. Could you solve it in-place? Remember that the board needs to be updated at the same time: You cannot update some cells first and then use their updated values to update other cells.
2. In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches the border of the array. How would you address these problems?

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因为要实现就地算法，所以必须要在更新数组的时候同时保持原有信息不丢失，相当于一个编码问题。本题只有四种情况：从 0 变为 0，从 0 变为 1，从 1 变为 1，从 1 变为 0。如果把这四种情况分别标记为0，2，1，3，那么统计周围八格有多少存活细胞的时候只需要看周围八个格子里面有多少个奇数即可。最后解码回来如果是1或2就是细胞存活，0或3就是细胞死亡。

 

对于优化和follow up问题，直接贴上大神的帖子https://segmentfault.com/a/1190000003819277。

 

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Java

class Solution {
    public void gameOfLife(int[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) return;
        int m = board.length, n = board[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                board[i][j] = status(board, i ,j);
            }
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == 1 || board[i][j] == 2) board[i][j] = 1;
                else board[i][j] = 0;
            }
        }
    }
    
    private int status(int[][] board, int i, int j) {
        boolean isAlive = board[i][j] % 2 == 1;
        int count = 0;
        int rowS = Math.max(i - 1, 0), colS = Math.max(j - 1, 0), rowE = Math.min(i + 1, board.length - 1), colE = Math.min(j + 1, board[0].length - 1);
        for (int p = rowS; p <= rowE; p++) {
            for (int q = colS; q <= colE; q++) {
                if (p == i && q == j) continue;
                if (board[p][q] % 2 == 1) count++;
            }
        }
        if (isAlive && count < 2) return 3;
        else if (isAlive && (count == 2 || count == 3)) return 1;
        else if (isAlive && count > 3) return 3;
        else if (!isAlive && count == 3) return 2;
        else return 0;
    }
}