LeetCode:Game of Life

Game of Life

Total Accepted: 20082  Total Submissions: 57678  Difficulty: Medium

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.

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思路：

细胞的转化情况有四种：

dead -> dead

live-> dead

dead-> live

live-> live

由于题目要求用in-place就地操作。因此在转化的过程中，即要体现出转化结果也要保留转化之前的状态。

由于转化前、后都是两种状态，因此可以用2个bit来表示：[bit2, bit1]

bit1：转化前状态；

bit2：转化后状态。

因此，上面的转化情况表示为：

00:dead <- dead

01:dead <- live

10:live <- dead

11:live <- live

由于bit2默认为0，因此只考虑10和11两种情况即可。

board[i][j] & 1：表示取bit1；

board[i][j] >> 1：表示取bit2。

java code:

public class Solution {
    public void gameOfLife(int[][] board) {
        
        if(board == null || board.length == 0) return;
        
        int rows = board.length;
        int cols = board[0].length;
        
        for(int i=0;i<rows;i++) {
            for(int j=0;j<cols;j++) {
                int lives = liveInNeighbors(board, rows, cols, i, j);
                
                // 只考虑第2位0->1和1->1的情况即可，因为第2位默认为0
                if(board[i][j]==0 && lives==3)
                    board[i][j] = 2;
                if(board[i][j]==1 && (lives==2 || lives==3))
                    board[i][j] = 3;
            }
        }
        
        for(int i=0;i<rows;i++) {
            for(int j=0;j<cols;j++) {
                board[i][j] >>= 1; // 取高位
            }
        }
    }
    
    private int liveInNeighbors(int board[][], int rows, int cols, int row, int col) {
        
        int lives = 0;
        
        for(int i = Math.max(0, row-1);i <= Math.min(rows-1, row+1);i++) {
            for(int j = Math.max(0, col-1);j <= Math.min(cols-1, col+1);j++) {
                lives += board[i][j] & 1; // 取低位
            }
        }
        lives -= board[row][col] & 1;
        return lives;
    }
}