LeetCode 289. Game of Life

分析

难度 中
来源 https://leetcode.com/problems/game-of-life/

思路

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

To get the current state, simply do board[i][j] & 1 To get the next
state, simply do board[i][j] >> 1

题目

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. 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?

解答

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){//横向 纵向均有不小于1
        int count=0;
        for(int row=Math.max(i-1,0);row<=Math.min(board.length-1,i+1);row++)//行
        {
            for(int col=Math.max(j-1,0);col<=Math.min(board[0].length-1,j+1);col++)//列
            {
                count+=board[row][col]&1;//board[row][col]==1时board[row][col]&1==1，board[row][col]==0时board[row][col]&1==0
            }
        }
        //对board[i][j]进行了一次统计
        return count-(board[i][j]&1);
    }
    public void gameOfLife(int[][] board) {//也可以使用Board的最高位存储当前状态，低位存储存活邻居数
        int rowNum=board.length;
        int colNum=board[0].length;
        if(rowNum==0||colNum==0)
            return;
        for(int i=0;i3)
                        board[i][j]=1;//当前1 下一步0。此时board[i][j]不需要改变
                    else //(liveNeighbor==2||liveNeighbor==3)
                        board[i][j]=3;//当前1 下一步1*/
                        board[i][j]=3;
                }else if(((board[i][j]&1)==0)&&(liveNeighbor==3))//当前0 下一步1。毕竟与上边的if不是独立的。
                    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