Cinte's Leetcode Record
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289. 生命游戏

Posted on Edited on In Disqus:
Symbols count in article: 1.2k Reading time ≈ 1 mins.

289. 生命游戏

0和1只需要int中的最低位就可以保存了，所以我们用int中的第二位来存储下一个状态的结果。

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class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        int m = board.size(), n = board[0].size();
        for(int i = 0; i < m; ++i)
        {
            for(int j = 0; j < n; ++j)
            {
                int aliveCell = 0;
                int cur = board[i][j];
                if(i > 0 && board[i - 1][j] & 1)
                    ++aliveCell;
                if(j > 0 && board[i][j - 1] & 1)
                    ++aliveCell;
                if(i < m - 1 && board[i + 1][j] & 1)
                    ++aliveCell;
                if(j < n - 1 && board[i][j + 1] & 1)
                    ++aliveCell;
                if(i > 0 && j > 0 && board[i - 1][j - 1] & 1)
                    ++aliveCell;
                if(i < m - 1 && j > 0 && board[i + 1][j - 1] & 1)
                    ++aliveCell;
                if(i < m - 1 && j < n - 1 && board[i + 1][j + 1] & 1)
                    ++aliveCell;
                if(i > 0 && j < n - 1 && board[i - 1][j + 1] & 1)
                    ++aliveCell;
                if(cur == 1 && (aliveCell < 2 || aliveCell > 3))
                    board[i][j] = 1;
                else if(cur == 0 && aliveCell == 3)
                    board[i][j] = 2;
                else
                    board[i][j] = board[i][j] ? 3 : 0;
            }
        }
        for(auto& line : board)
            for(auto& num : line)
                num >>= 1;
        /*
        for_each(board.begin(), board.end(), [](vector<int>& line){
            for_each(line.begin(), line.end(), [](int& num){
                num >>= 1;
            });
        });
        */
    }
};