289. 生命游戏

要求：细胞矩阵按规则更新https://leetcode-cn.com/problems/game-of-life/
思路：
法一：用额外矩阵

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        int neighbors[3] = {0, 1, -1};
        int rows = board.size();
        int cols = board[0].size();
        // 创建复制数组 copyBoard
        vector<vector<int> >copyBoard(rows, vector<int>(cols, 0));
        // 从原数组复制一份到 copyBoard 中
        for (int row = 0; row < rows; row++) 
            for (int col = 0; col < cols; col++) 
                copyBoard[row][col] = board[row][col];
        // 遍历面板每一个格子里的细胞
        for (int row = 0; row < rows; row++) {
            for (int col = 0; col < cols; col++) {
                // 对于每一个细胞统计其八个相邻位置里的活细胞数量
                int liveNeighbors = 0;
                for (int i = 0; i < 3; i++) {
                    for (int j = 0; j < 3; j++) {
                        if (!(neighbors[i] == 0 && neighbors[j] == 0)) {
                            int r = (row + neighbors[i]);
                            int c = (col + neighbors[j]);
                            // 查看相邻的细胞是否是活细胞
                            if ((r < rows && r >= 0) && (c < cols && c >= 0) && (copyBoard[r][c] == 1)) {
                                liveNeighbors += 1;
                            }
                        }
                    }
                }
                // 规则 1 或规则 3      
                if ((copyBoard[row][col] == 1) && (liveNeighbors < 2 || liveNeighbors > 3)) 
                    board[row][col] = 0;
                // 规则 4
                if (copyBoard[row][col] == 0 && liveNeighbors == 3) 
                    board[row][col] = 1;
            }
        }
    }
};

法二：不用多余空间，更新成其他数字即可

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        int neighbors[3] = {0, 1, -1};
        int rows = board.size();
        int cols = board[0].size();
        // 遍历面板每一个格子里的细胞
        for (int row = 0; row < rows; row++) {
            for (int col = 0; col < cols; col++) {
                // 对于每一个细胞统计其八个相邻位置里的活细胞数量
                int liveNeighbors = 0;
                for (int i = 0; i < 3; i++) {
                    for (int j = 0; j < 3; j++) {
                        if (!(neighbors[i] == 0 && neighbors[j] == 0)) {
                            // 相邻位置的坐标
                            int r = (row + neighbors[i]);
                            int c = (col + neighbors[j]);
                            // 查看相邻的细胞是否是活细胞
                            if ((r < rows && r >= 0) && (c < cols && c >= 0) && (abs(board[r][c]) == 1)) {
                                liveNeighbors += 1;
                            }
                        }
                    }
                }
                // 规则 1 或规则 3 
                if ((board[row][col] == 1) && (liveNeighbors < 2 || liveNeighbors > 3)) 
                    // -1 代表这个细胞过去是活的现在死了
                    board[row][col] = -1;
                // 规则 4
                if (board[row][col] == 0 && liveNeighbors == 3) 
                    // 2 代表这个细胞过去是死的现在活了
                    board[row][col] = 2;
            }
        }
        // 遍历 board 得到一次更新后的状态
        for (int row = 0; row < rows; row++) {
            for (int col = 0; col < cols; col++) {
                if (board[row][col] > 0) {
                    board[row][col] = 1;
                } else {
                    board[row][col] = 0;
                }
            }
        }
    }
};