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?

1.用額外空間的。先根據每個live向四面八方做貢獻，給周圍格子live計數++。從而先得到整個棋盤每個點的四周live計數矩陣。之後根據該矩陣和條件更新狀態。

2.利用2位記錄狀態。00,01,10,11，低位表示當前狀態，高位表示下輪狀態。這樣計數的時候只看board[i][j] & 1即可。就算把1改2（下輪要變0）也不會影響後續計數。

實現1：

class Solution {
    public void gameOfLife(int[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) {
            return;
        }
        int[][] count = new int[board.length][board[0].length];
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                count(board, i, j, count);
            }
        }
        
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                int cnt = count[i][j];
                if (board[i][j] == 0 && cnt == 3) {
                    board[i][j] = 1;
                } else if (board[i][j] == 1 && (cnt < 2 || cnt > 3)) {
                    board[i][j] = 0;
                }
            }
        }
    }

    private void count(int[][] board, int x, int y, int[][] count) {
        if (board[x][y] == 0) {
            return;
        }
        int[] dx = {0, 1, 0, -1, 1, 1, -1, -1};
        int[] dy = {1, 0, -1, 0, 1, -1, 1, -1};
        for (int i = 0; i < 8; i++) {
            int newX = x + dx[i];
            int newY = y + dy[i];
            if (isInBound(board, newX, newY)) {
                count[newX][newY]++;
            }
        }
    }

    private boolean isInBound(int[][] board, int x, int y) {
        return x >= 0 && x < board.length && y >= 0 && y < board[0].length;
    }
}

2.實現2

class Solution {
    public void gameOfLife(int[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) {
            return;
        }
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                int cnt = count(board, i, j);
                if (board[i][j] == 0 && cnt == 3) {
                    board[i][j] = 2;
                } else if (board[i][j] == 1 && (cnt < 2 || cnt > 3)) {
                    board[i][j] = 5;
                } else if (board[i][j] == 1) {
                    board[i][j] = 3;
                }
            }
        }

        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                board[i][j] = ((board[i][j] >> 1) & 1);
            }
        }
    }

    private int count(int[][] board, int x, int y) {
        int[] dx = {0, 1, 0, -1, 1, 1, -1, -1};
        int[] dy = {1, 0, -1, 0, 1, -1, 1, -1};
        int cnt = 0;
        for (int i = 0; i < 8; i++) {
            int newX = x + dx[i];
            int newY = y + dy[i];
            if (isInBound(board, newX, newY) && (board[newX][newY] & 1) == 1) {
                cnt++;
            }
        }
        return cnt;
    }

    private boolean isInBound(int[][] board, int x, int y) {
        return x >= 0 && x < board.length && y >= 0 && y < board[0].length;
    }
}

leetcode289- Game of Life- medium