題目來源
一道狀態(tài)轉(zhuǎn)移的題目抵知，題目比較長(zhǎng)…然后實(shí)際做起來并不難堤魁。我一開始做利用了mn的空間耕魄，然后處理起來比較冗長(zhǎng)晨川。
代碼如下：

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;
            }
    }
};

然后原來感覺這種題不值得寫東西的证九，但是看了大神們簡(jiǎn)潔的代碼删豺，覺得還是得寫一下學(xué)習(xí)一下，以后這種隔壁八塊愧怜，判斷是否越界的呀页，可以這么寫，然后判斷是哪個(gè)狀態(tài)的叫搁，直接用與或這些操作來判斷赔桌，簡(jiǎn)潔的很供炎，厲害渴逻，實(shí)在是佩服。見代碼：

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;
    }
};