Leetcode 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?

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解题思路：

位运算（bit manipulation）

由于细胞只有两种状态0和1，因此可以使用二进制来表示细胞的生存状态

更新细胞状态时，将细胞的下一个状态用高位进行存储

全部更新完毕后，将细胞的状态右移一位

To solve it in place, we use 2 bits to store 2 states:

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

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

In the beginning, every 2nd state is 0; when next becomes alive change 2nd bit to 1:

• live -> die: nbs < 2 || nbs > 3 (we don't need to care!)
• live -> live: nbs >= 2 && nbs <= 3
• dead -> live: nbs == 3

To get this state, we simple do:

board[i][j] & 1

To get next state, we simple do:

board[i][j] >> 1

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Java code:

public class Solution {
    public void gameOfLife(int[][] board) {
        if(board == null || board.length == 0) {
            return;
        }
        int m = board.length, n = board[0].length;
        // In the beginning, every 2nd bit is 0;
        // So we only need to care about when 2nd bit will become 1.
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                int lives = liveNeighbors(board, m, n, i , j);
                if((board[i][j] & 1) == 1 && (lives >= 2 && lives <= 3)) {
                    board[i][j] = 3;  //Make the 2nd bit 1: 01----> 11
                }else if ((board[i][j] & 1) == 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
            }
        }
    }
    
    private int liveNeighbors(int[][]board, int m, int n, int i , int j){
        int lives = 0;
        int[][] d= {{1,-1},{1,0},{1,1},{0,-1},{0,1},{-1,-1},{-1,0},{-1,1}};
        for(int k = 0; k < 8; k++) {
            int x = d[k][0] + i;
            int y = d[k][1] + j;
            if( x < 0 || x >= m || y < 0 || y >= n){
                continue;
            }
            if((board[x][y] & 1) == 1) {
                lives++;
            }
        }
        return lives;
    }
}

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Reference:

1. https://leetcode.com/discuss/68352/easiest-java-solution-with-explanation

2. https://leetcode.com/discuss/62038/c-ac-code-o-1-space-o-mn-time

3. http://bookshadow.com/weblog/2015/10/04/leetcode-game-life/