LeetCode *** 289. Game of Life

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本文探讨了基于Wikipedia文章的“游戏生命”细胞自动机的实现，详细介绍了其核心规则，并提供了一种通过位操作解决游戏生命状态更新问题的方法。讨论了在原地解决此问题的挑战以及如何处理无限板的边界问题。

题目：

According to the Wikipedia's article: "The Game of Life, also known simply asLife, 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 statelive (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:

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.
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?

分析：

位操作即可。

代码：

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