【Leetcode】289. Game of Life

Description

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):

Any live cell with fewer than two live neighbors dies, as if caused by under-population.
Any live cell with two or three live neighbors lives on to the next generation.
Any live cell with more than three live neighbors dies, as if by over-population..
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.

Solution

public class Solution {
    public void gameOfLife(int[][] board) {
            /*
        根据题意,是这么几个状态:
        第一位是初始状态,第二位是后续状态
        0.00 初始挂,后续挂
        1.01 初始活,后续挂
        2.10 初始挂,后续活
        3.11 初始活,后续活
        首先要统计周围的存活个数,然后根据规则来判断:
        1.周围存活小于2个,活的会挂
        2.周围存活有2个或者3个,活的会活
        3.周围存活超过3个,活的会挂
        4.周围3个活的, 挂的会活
         */
        int m = board.length;
        int n = board[0].length;
        if (board == null || (m == 0 && n == 0)) {
            return;
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int lives = countLive(board, i, j, m, n);
                //现在board要么是0,要么是1
                //然后board变为10,要么是11才算存活下来
                if (board[i][j] == 1 && lives >= 2 && lives <= 3) {
                    board[i][j] = 3;
                }
                if (board[i][j] == 0 && lives == 3) {
                    board[i][j] = 2;
                }
            }
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                board[i][j] >>= 1;
            }
        }
    }
    
    public int countLive(int[][] board, int i , int j, int m, int n) {
        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;
    }
}

Analyse

注释已经写得很清楚了，这个解法真是清新易懂。