Game of Life

2017-04-20  本文已影响16人  我叫胆小我喜欢小心

题目来源
一道状态转移的题目,题目比较长…然后实际做起来并不难。我一开始做利用了mn的空间,然后处理起来比较冗长。
代码如下:

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        int rows = board.size();
        if (rows == 0)
            return;
        int cols = board[0].size();
        vector<vector<int>> neighbors(rows, vector<int>(cols, 0));
        for (int row=0; row<rows; row++)
            for (int col=0; col<cols; col++) {
                if (board[row][col] == 1) {
                    if (row > 0)
                        neighbors[row-1][col]++;
                    if (row > 0 && col > 0)
                        neighbors[row-1][col-1]++;
                    if (col > 0)
                        neighbors[row][col-1]++;
                    if (row < rows - 1 && col > 0)
                        neighbors[row+1][col-1]++;
                    if (row < rows - 1)
                        neighbors[row+1][col]++;
                    if (row < rows - 1 && col < cols - 1)
                        neighbors[row+1][col+1]++;
                    if (col < cols - 1)
                        neighbors[row][col+1]++;
                    if (row > 0 && col < cols - 1)
                        neighbors[row-1][col+1]++;
                }
            }
        for (int row=0; row<rows; row++)
            for (int col=0; col<cols; col++) {
                if (board[row][col] == 1 && (neighbors[row][col] < 2 || neighbors[row][col] > 3))
                    board[row][col] = 0;
                if (board[row][col] == 0 && neighbors[row][col] == 3)
                    board[row][col] = 1;
            }
    }
};

然后原来感觉这种题不值得写东西的,但是看了大神们简洁的代码,觉得还是得写一下学习一下,以后这种隔壁八块,判断是否越界的,可以这么写,然后判断是哪个状态的,直接用与或这些操作来判断,简洁的很,厉害,实在是佩服。见代码:

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        int rows = board.size(), cols = (rows == 0) ? 0 : board[0].size();
        for (int row=0; row<rows; row++)
            for (int col=0; col<cols; col++) {
                int cnt = 0;
                for (int i=max(0, row-1); i<=min(rows-1, row+1); i++)
                    for (int j=max(0, col-1); j<=min(cols-1, col+1); j++)
                        cnt += board[i][j] & 1;
                if (cnt == 3 || cnt - board[row][col] == 3)
                    board[row][col] |= 2;
            }
        for (int row=0; row<rows; row++)
            for (int col=0; col<cols; col++)
                board[row][col] >>= 1;
    }
};
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