289. Game of Life

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?

一刷
题解：

[2nd bit, 1st bit] = [next state, current state]

- 00  dead (next) <- dead (current)
- 01  dead (next) <- live (current)  
- 10  live (next) <- dead (current)  
- 11  live (next) <- live (current) 

对每一个cell，计算出周围活着的cell数目，并且修改2nd bit
最后全部往右移一位, 得到next status

class Solution {
    public void gameOfLife(int[][] board) {
        if (board == null || board.length == 0) return;
        int m = board.length, n = board[0].length;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int lives = liveNeighbors(board, m, n, i, j);

                // In the beginning, every 2nd bit is 0;
                // So we only need to care about when will the 2nd bit become 1.
                if (board[i][j] == 1 && lives >= 2 && lives <= 3) {  
                    board[i][j] = 3; // Make the 2nd bit 1: 01 ---> 11
                }
                if (board[i][j] == 0 && lives == 3) {
                    board[i][j] = 2; // Make the 2nd bit 1: 00 ---> 10
                }
            }
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                board[i][j] >>= 1;  // Get the 2nd state.
            }
        }
    }

    public int liveNeighbors(int[][] board, int m, int n, int i, int j) {
        int lives = 0;
        for (int x = Math.max(i - 1, 0); x <= Math.min(i + 1, m - 1); x++) {
            for (int y = Math.max(j - 1, 0); y <= Math.min(j + 1, n - 1); y++) {
                lives += board[x][y] & 1;
            }
        }
        lives -= board[i][j] & 1;
        return lives;
    }

}