[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?

Credits:

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

给定一个由0，1组成的矩阵，每一个元素表示一个细胞的存活，1存活，0死亡，其中下一次更新每个细胞的存活由上、下、左、右、左上、左下、右上、右下，八个细胞决定，存活规则如下：

     1、周围存活细胞小于2个，则死亡。

    2、周围存活细胞2个或3个，则存活到下一次更新。

    3、周围存活细胞3个以上，则死亡。

    4、周围存活细胞3个的死细胞，将会成为一个活细胞。

AC代码：

class Solution{public:    void gameOfLife(vector<vector<int>>& board)    {        int rows=board.size();        if(rows==0)            return ;        int colums=board[0].size();        if(colums==0)            return ;        for(int i=0; i<rows; ++i)        {            for(int j=0; j<colums; ++j)            {                int sum=getNeighbor(board,rows,colums,i,j);                if(sum==2)                    continue;                else if(sum==3)                    board[i][j]=board[i][j]==0?3:1;                else                    board[i][j]=board[i][j]==1?2:0;            }        }        for(int i=0;i<rows;++i)        {            for(int j=0;j<colums;++j)                board[i][j]%=2;        }    }private:    int getNeighbor(vector<vector<int>>& board,int rows,int colums,int x,int y)    {        int sum=0;        for(int i=x-1; i<x+2; ++i)        {            for(int j=y-1; j<y+2; ++j)            {                if(i==x&&j==y)                    continue;                if(i>=0&&i<rows&&j>=0&&j<colums&&(board[i][j]==1||board[i][j]==2))                    ++sum;            }        }        return sum;    }};

其他leetcode题目AC代码：https://github.com/PoughER