LeetCode 289 - Game of Life

Title Description

289. Game of Life

solution:

Here to talk about bit-manipulation Kazakhstan:

An int accounted for 32, 0, 1, but only one, we can make use of all the rest of the digits do mark. Specifically, by using only the second can,

. 1 rule : the original 01, the number of viable cells is less than about 2, a right shift is 0
rule 2 : the original 01, the number of viable cells is equal to about 2, or 3,01 or 10 is 11, one is right. 1
rule 3 : the original is 01, the number of surrounding cell survival greater than 3, the right one is 0
rule 4 : the original is 00, equal to the number of cell survival around 3.00 or 10 is 10, is a right one

Really was a god operation! !

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        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<board.size();i++)
        {
            for(int j=0;j<board[0].size();j++)
            {
                int sum = 0;
                for(int k=0;k<8;k++)
                {
                    int nx = i + dx[k];
                    int ny = j + dy[k];
                    if((nx>=0 && nx<board.size()) && (ny>=0 && ny<board[0].size()))
                        sum += (board[nx][ny]&1); // 只累加最低位
                }
                if(board[i][j]==1)
                {
                    if(sum==2 || sum==3) board[i][j] |= 2; // 使用第二个bit标记是否存活
                }
                else
                {
                    if(sum==3) board[i][j] |= 2;
                }   
            }
        }
        for(int i=0;i<board.size();i++)
        {
            for(int j=0;j<board[0].size();j++)
                board[i][j] >>= 1;
        }
    }
};