leetcode 289. Game of Life

leetcode 289. Game of Life

谷歌的面试题果然很开眼界，比如这题，用两bit来表示前后两次的状态，很巧妙。

public class Solution {
    public void gameOfLife(int[][] board) {
        if(board==null||board.length==0) return;
        int m = board.length;
        int n = board[0].length;
        // if(n==0) return;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                int lives = getlivesaround(board,m,n,i,j);
                if(board[i][j]==1&&(lives==2||lives==3)){
                    board[i][j] = 3; // 11<-01
                }
                if(board[i][j]==0&&lives==3){
                    board[i][j] = 2;  // 10<-00
                }
            }
        }
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                board[i][j] = board[i][j]>>1;
            }
        }
        
    }
    
    private int getlivesaround(int[][] board, int m, int n, int i, int j){
        int lives = 0;
        for(int h = Math.max(i-1,0);h <= Math.min(i+1, m-1);h++){
            for(int k = Math.max(j-1,0);k <= Math.min(j+1, n-1);k++){
                lives += board[h][k]&1;
            }
        }
        lives -= board[i][j]&1;
        return lives;
    }
}