[289]Game of Life

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本文探讨了由英国数学家约翰·康威于1970年发明的游戏生命（Life）的实现，涉及二维矩阵中细胞的状态转换，遵循特定规则进行下一代状态的计算。文章深入分析了细胞状态变化的四种情况，并提供了在原地更新矩阵的方法，同时考虑了矩阵边界条件的处理。

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【题目描述】

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:

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?

【思路】

根据题意，cell有四种变化状态dead->dead,dead->live,live->dead,live->live,因为要保证所有cell同时更新，所以利用十位，给这种状态分别赋值00,10,01,11.这样我们仍可以利用其个位来帮助该cell周围的cell更新状态，又可以根据其十位判断该cell更新后的状态。

【代码】

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        int m=board.size();
        int n=board[0].size();
        if(m==0||n==0) return;
        int c=0;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(board[i][j]==1){
                    c=getlivecell(board,i,j);
                    if(c==2||c==3) board[i][j]+=10;
                }
                else{
                   c=getlivecell(board,i,j);
                   if(c==3) board[i][j]+=10;
                }
            }
        }
         for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                board[i][j]/=10;
            }
        }
    }
    int getlivecell(vector<vector<int>>& board,int a,int b){
        int m=board.size();
        int n=board[0].size();
        int cnt=0;
        for(int i=a-1;i<=a+1;i++){
            for(int j=b-1;j<=b+1;j++){
                if(i<0||j<0||i>m-1||j>n-1||(i==a&&j==b)) continue;
                if(board[i][j]%10==1) cnt++;
            }
        } 
        return cnt;
    }
};