Game of Life 生命游戏(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."

  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?
    3.

题目大意：
根据维基百科的文章：“生命游戏，也被简称为生命，是一款由英国数学家约翰·霍顿康威于1970年设计的细胞自动机。”

给定一个m * n的细胞隔板，每一个细胞拥有一个初始状态：存活(1)或者死亡(0)。每一个细胞与其周围的8个邻居细胞（水平，竖直，对角线）发生交互，依据如下四条规则（摘自维基百科）：

任何相邻存活细胞数小于2个的存活细胞都会死亡，模拟人口不足。

任何相邻存活细胞数为2个或者3个的存活细胞会存活到下一代。

任何相邻存活细胞数大于3个的存活细胞都会死亡，模拟人口过载。

任何相邻存活细胞数等于3个的死亡细胞都会成为一个存活细胞，模拟繁殖。

编写函数，根据隔板的当前状态，计算其下一个状态（一次更新之后）

进一步思考：

你可以就地完成题目吗？记住隔板需要同时更新：你不能先更新某些细胞然后再以其变更后的值来更新其他细胞。

在这个问题中，我们使用2维数组表示隔板。原则上，隔板是无穷的，这可能导致一些边界问题。你怎么处理边界问题？

解题思路

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)
  • In the beginning, every cell is either 00 or 01.
  • Notice that 1st state is independent of 2nd state.
  • Imagine all cells are instantly changing from the 1st to the 2nd state, at the same time.
  • Let's count # of neighbors from 1st state and set 2nd state bit.
  • Since every 2nd state is by default dead, no need to consider transition 01 -> 00.
  • In the end, delete every cell's 1st state by doing >> 1.
For each cell's 1st bit, check the 8 pixels around itself, and set the cell's 2nd bit.

Transition 01 -> 11: when board == 1 and lives >= 2 && lives <= 3.
Transition 00 -> 10: when board == 0 and lives == 3.

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

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

Hope this helps!

public void gameOfLife(int[][] board) {
    if(board == null || board.length == 0) return;
    int m = board.length, n = board[0].length;

    for(int i = 0; i < m; i++) {
        for(int j = 0; j < n; j++) {
            int lives = liveNeighbors(board, m, n, i, j);

            // In the beginning, every 2nd bit is 0;
            // So we only need to care about when the 2nd bit will become 1.
            if(board[i][j] == 1 && lives >= 2 && lives <= 3) {  
                board[i][j] = 3; // Make the 2nd bit 1: 01 ---> 11
            }
            if(board[i][j] == 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.
        }
    }
}

public int liveNeighbors(int[][] board, int m, int n, int i, int j) {
    int lives = 0;
    for(int x = Math.max(i - 1, 0); x <= Math.min(i + 1, m - 1); x++) {
        for(int y = Math.max(j - 1, 0); y <= Math.min(j + 1, n - 1); y++) {
            lives += board[x][y] & 1;
        }
    }
    lives -= board[i][j] & 1;
    return lives;
}

Reference

1 neverlandly: Game of Life

2 Easiest JAVA solution with explanation

results matching ""

    No results matching ""