LeetCode Game of Life(位操作)

最新推荐文章于 2020-11-19 00:14:42 发布

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本文介绍了一个基于二维数组实现的生命游戏算法。该算法遵循英国数学家约翰·康威于1970年提出的细胞自动机原理，通过四个核心规则来更新每个单元格的状态。文章详细解释了如何在不改变原有状态的情况下计算出下一代的状态，并提供了具体的实现代码。

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?

题意：给出一个二维数组，数组中的元素为0或者1两种状态，数组中元素状态会根据周围的8个方格的状态变化，求其变化后的状态。

思路：因为数组中的元素是0或者1，在变化后的状态放在第2位上，然后将所有元素向右移1位

代码如下：

class Solution
{
    public void gameOfLife(int[][] board)
    {
        if (0 == board.length) return;

        int row = board.length;
        int col = board[0].length;
        int[] dx = {-1, 0, 1, -1, 1, -1, 0, 1};
        int[] dy = {-1, -1, -1, 0, 0, 1, 1, 1};

        for (int i = 0; i < row; i++)
        {
            for (int j = 0; j < col; j++)
            {
                int live = 0;
                for (int k = 0; k < 8; k++)
                {
                    int rx = i + dx[k];
                    int ry = j + dy[k];
                    if (rx < 0 || rx >= row || ry < 0 || ry >= col) continue;
                    live += board[rx][ry] & 1;
                }
                if (3 == live + board[i][j] || 3 == live)
                {
                    board[i][j] |= 2;
                }
            }
        }

        for (int i = 0; i < row; i++)
        {
            for (int j = 0; j < col; j++)
            {
                board[i][j] >>= 1;
            }
        }
    }
}