LeetCode 289. Game of Life

ぶんせき
難易度のソースhttps://leetcode.com/problems/game-of-life/
構想
To solve it in place, we use 2 bits to store 2 states: [2nd bit, 1st bit] = [next state, current state]

00 dead (next)

01 dead (next)

10 live (next)

11 live (next)

To get the current state, simply do board[i][j] & 1 To get the next state, simply do board[i][j] >> 1
タイトル
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. The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously. Example: Input:

[ [0,1,0], [0,0,1], [1,1,1], [0,0,0] ]
Output:
[ [0,0,0], [1,0,1], [0,1,1], [0,1,0] ]
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?

に答える
Runtime: 1 ms, faster than 100.00% of Java online submissions for Game of Life. Memory Usage: 37.3 MB, less than 39.53% of Java online submissions for Game of Life.

package LeetCode;

import java.util.Arrays;

public class L289_GameOfLife {
    private int liveNeighbors(int[][] board,int i,int j){//          1
        int count=0;
        for(int row=Math.max(i-1,0);row<=Math.min(board.length-1,i+1);row++)// 
        {
            for(int col=Math.max(j-1,0);col<=Math.min(board[0].length-1,j+1);col++)// 
            {
                count+=board[row][col]&1;//board[row][col]==1 board[row][col]&1==1，board[row][col]==0 board[row][col]&1==0
            }
        }
        // board[i][j]       
        return count-(board[i][j]&1);
    }
    public void gameOfLife(int[][] board) {//     Board          ，         
        int rowNum=board.length;
        int colNum=board[0].length;
        if(rowNum==0||colNum==0)
            return;
        for(int i=0;i3)
                        board[i][j]=1;//  1    0。  board[i][j]     
                    else //(liveNeighbor==2||liveNeighbor==3)
                        board[i][j]=3;//  1    1*/
                        board[i][j]=3;
                }else if(((board[i][j]&1)==0)&&(liveNeighbor==3))//  0    1。      if     。
                    board[i][j]=2;
            }
        }
        for(int i=0;i>1;
            }
        }
    }

    public static void main(String[] args){
        L289_GameOfLife l289=new L289_GameOfLife();
        int[][] input= {{0,1,0},
                        {0,0,1},
                        {1,1,1},
                        {0,0,0}};
        l289.gameOfLife(input);
        System.out.println();
        for(int i=0;i