LeetCode-Game of Life

Posted 2020-08-08 LiBlog

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.

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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

 Analysis:

We need a way to encode a position‘s good neighbors and also its existing status:

Dead: - (# of live neighbors) -1

Live: (# of live neighbors) +1

Note: the -1 and +1 is used to seperate the states with 0 live neighbors.

 

Another way is to calculate the next status in-place and use 2 bits coding,

00: dead but will live

01: dead and will die

10: live but will die

11: live and will live

 

Solution:

public class Solution {
    public void gameOfLife(int[][] board) {
    if(board == null || board.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++) {
            int lives = liveNeighbors(board, m, n, i, j);

            // In the beginning, every 2nd bit is 0;
            // So we only need to care about when the 2nd bit will become 1.
            if(board[i][j] == 1 && lives >= 2 && lives <= 3) {  
                board[i][j] = 3; // Make the 2nd bit 1: 01 ---> 11
            }
            if(board[i][j] == 0 && lives == 3) {
                board[i][j] = 2; // Make the 2nd bit 1: 00 ---> 10
            }
        }
    }

    for(int i = 0; i < m; i++) {
        for(int j = 0; j < n; j++) {
            board[i][j] >>= 1;  // Get the 2nd state.
        }
    }
}

public int liveNeighbors(int[][] board, int m, int n, int i, int j) {
    int lives = 0;
    for(int x = Math.max(i - 1, 0); x <= Math.min(i + 1, m - 1); x++) {
        for(int y = Math.max(j - 1, 0); y <= Math.min(j + 1, n - 1); y++) {
            lives += board[x][y] & 1;
        }
    }
    lives -= board[i][j] & 1;
    return lives;
}
}