LeetCode #289 - Game of Life

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):

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. The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously.

Example:

Input: 
[
  [0,1,0],
  [0,0,1],
  [1,1,1],
  [0,0,0]
]
Output: 
[
  [0,0,0],
  [1,0,1],
  [0,1,1],
  [0,1,0]
]

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?

按照生命游戏给定的规则对矩阵进行迭代，要求直接修改数组，但是实际上所有元素的更新是同步完成的，所以需要解决一个问题就是即使一个元素值已经更新过，还是必须用之前的值来更新其他元素。解决方法是利用两位编码也就是0~3四个数来表示，上一次的值和当前值的四种可能情况，从而可以根据之前的值更新其他点，并根据当前值更新下一次的状态。

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