Leetcode 289. Game of Life

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?

细胞自动机下一个状态的计算。状态更新是同时的，不能一个细胞一个细胞地更新。要求在原始数组上直接操作。

既然不能开新的数组，就只能通过原始数组记录了，考虑到0和1只占用最后一位，那么可以用其他bit位存储更新后的状态。

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        int dir[8][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
        int n = board.size();
        if(!n) return;
        int m = board[0].size();
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                int cnt = 0;
                for(int k = 0; k < 8 ; k++)
                {
                    int x = i + dir[k][0], y = j + dir[k][1];
                    if(x < 0 || x >= n || y < 0 || y >= m) continue;
                    if(board[x][y] & 1) cnt++;
                }
                if(cnt == 2) board[i][j] |= (board[i][j] & 1) << 1;
                else if(cnt == 3) board[i][j] |= 1 << 1;
            }
        }
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
                board[i][j] >>= 1;
    }
};