Game of Life

题目：

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.

分析：

因为要原地变换，所以可以借助状态机转换；

如果原节点为1，而下个状态要die了，就将其置为2；

如果源节点为0，而下个状态要活了，就将其置为3；

先遍历一边二维数组；

如果当前节点为1：

如果其邻居中1或者2的个数 小于2 或者大于3，则其将要死了，将其值变为2；

否则，如果当前节点为0：

如果邻居中1或者2的个数为3，则其将要活了，将其值变为3；

再遍历一边数组，对于每个点，value = value % 2;

 代码如下：

public class Solution {    public void gameOfLife(int[][] board) {        int m = board.length;        int n = board[0].length;        int dx[] = {-1, -1, -1, 0, 1, 1, 1, 0};        int dy[] = {-1, 0, 1, 1, 1, 0, -1, -1};        for (int i = 0; i < m; ++i) {            for (int j = 0; j < n; ++j) {                int cnt = 0;                for (int k = 0; k < 8; ++k) {                    int x = i + dx[k], y = j + dy[k];                    if (x >= 0 && x < m && y >= 0 && y < n && (board[x][y] == 1 || board[x][y] == 2)) {                        ++cnt;                    }                }                if (board[i][j] == 1 && (cnt < 2 || cnt > 3)) board[i][j] = 2;                else if (board[i][j] == 0 && cnt == 3) board[i][j] = 3;            }        }        for (int i = 0; i < m; ++i) {            for (int j = 0; j < n; ++j) {                board[i][j] %= 2;            }        }    }}