289. Game of Life

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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?

解法1：

这题的关键点在于要想到用状态机来存储过去的历史。如果一位上仅用0,1表示的话，那么只能表示两个状态。如果我们用2 bits（也就是0,1,2,3）来表示的话可以表示4个状态。
这里我们有
0 -> 0 : 0
1 -> 1 : 1
1 -> 0 : 2
0 -> 1 : 3
然后逐个扫描，根据规则update每一个元素。要注意的是判断live cell的时候要check两个数值，一个是1，一个是2。这个地方很容易错。
最后我们对结果数组对2取对数就可以了。
C++

1

Java

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public class Solution {
    public void gameOfLife(int[][] board) {
        if (board.length == 0 || board[0].length == 0) {
            return;
        }
        int m = board.length, n = board[0].length;
        // define directions
        int[][] directions = new int[][]{{0,-1},{-1,-1},{-1,0},{-1,1},{0,1},{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 < directions.length; k++) {
                    int row = i + directions[k][0];
                    int col = j + directions[k][1];
                    if (row >= 0 && row < m && col >= 0 && col < n && (board[row][col] == 1 || board[row][col] == 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;
            }
        }
        return;
    }
}