[leetcode] 289. Game of Life

最新推荐文章于 2025-08-10 20:50:18 发布

转载最新推荐文章于 2025-08-10 20:50:18 发布 · 191 阅读

文章标签：

#leetcode #medium

leetcode 专栏收录该内容

179 篇文章

订阅专栏

本文介绍了一个经典的细胞自动机游戏——康威的生命游戏，并提供了一种使用C++实现的in-place算法来更新游戏板状态。该算法巧妙地利用了每个单元格的两个比特位来编码其历史和未来状态。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

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?

解法一：

这道题要用in-place的做法，那就要考虑把cell的历史状态编码到新的状态中。很容易想到用2 bit就可以表达。

00: die -> die

01: live -> live

10: live -> die

11: die -> live

这里01和11调转了一下，因为初始的状态是1，如果用1表示die->live，就分不清他是被复活的还是原始就是live的。详细见code。

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        
        int dx[] = {-1,0 , 1,-1, 1,-1, 0, 1};
        int dy[] = {-1,-1,-1, 0, 0, 1, 1, 1};
        
        
        int rows = board.size();
        int cols = board[0].size();
        for(int i = 0; i< rows; i++){
            for(int j=0; j< cols; j++){
                int cnt = 0;
                for(int k=0; k<8; k++){
                    int x = i+dx[k], y = j+dy[k];
                    if(x>=0&&x<rows&&y>=0&&y<cols&&(board[x][y]==2||board[x][y]==1)) cnt++;
                }
                if(board[i][j]==1&&(cnt<2||cnt>3)) board[i][j]=2;
                else if (board[i][j]==0&&cnt==3) board[i][j]=3;
            }
        }
        
        for(int i=0; i<rows; i++)
            for(int j=0; j<cols; j++)
                board[i][j] %= 2;
        
        
    }
};