289. Game of Life

问题描述：

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?

解题思路：

可以看到是根据矩阵元素根据它周围的9个元素的状态来判断下一代的状态：

简单的罗列出两种可能：1->0; 0->1

即使1要变0，那也是下一代的事情，现在它周围的邻居还认为它活着。

所以我们将1->0用2 来表示。这样检查的时候可以检查是否>0

0->1用-1表示，不影响本局其它邻居的遍历

代码：

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