Game of Life

According to the Wikipedia's article: "The Game of Life, also known simply as Life, is a cellular automaton devised by the British mathematician John Horton Conway in 1970."

Given a board with m by n cells, each cell has an initial state live (1) or dead (0). Each cell interacts with its eight neighbors (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article):

 

1. Any live cell with fewer than two live neighbors dies, as if caused by under-population.
2. Any live cell with two or three live neighbors lives on to the next generation.
3. Any live cell with more than three live neighbors dies, as if by over-population..
4. Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.

 

Write a function to compute the next state (after one update) of the board given its current state.

Follow up: 

1. Could you solve it in-place? Remember that the board needs to be updated at the same time: You cannot update some cells first and then use their updated values to update other cells.
2. In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches the border of the array. How would you address these problems?

解题思路：因为要求in-place,而状态转换最多也就四种，且只要记录两种即0->1,1->0这两种即可。于是可以现将board的值置为-1，-2，分别赋予两种状态转换的含义，最后再把负数换成本要转换的0，1状态即可。

class Solution {
public:
    //-1: 1->0 -2:0->1
    void countNeighbors(vector<vector<int>>& board, int x, int y, int& nLive) {
        int n = board.size(), m = board[0].size();
        int dir[8][2] = {
            {-1, -1}, {-1, 0}, {-1, 1},
            {0, -1}, {0, 1},
            {1, -1}, {1, 0}, {1,1}
        };
        for(int i = 0; i < 8; ++i) {
            int xx = x + dir[i][0];
            int yy = y + dir[i][1];
            if(xx >= 0 && xx < n && yy >= 0 && yy < m) {
                if(board[xx][yy] == 1 || board[xx][yy] == -1) nLive++;
            }
        }
    }
    void gameOfLife(vector<vector<int>>& board) {
        if(board.empty()) return;
        int n = board.size(), m = board[0].size(), nLive;
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < m; ++j) {
                nLive=0;
                countNeighbors(board, i, j, nLive);
                if(board[i][j] == 1) {
                    if(nLive < 2 || nLive > 3) board[i][j] = -1;
                }
                else {
                    if(nLive == 3) board[i][j] = -2;
                }
            }
        }
        for(int i = 0; i < n; ++i) 
            for(int j = 0; j < m; ++j) {
                if(board[i][j] == -1) board[i][j] = 0;
                else if(board[i][j] == -2) board[i][j] = 1;
            }
    }
};

看了discuss里的，用低一位表示初始状态，高一位表示改变之后的状态，这样最后只要去掉初始状态就能得到最终状态。ps：八个方向搜索可以借鉴下：

public void gameOfLife(int[][] board) {
    if (board == null || board.length == 0) return;
    int m = board.length, n = board[0].length;

    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            int lives = liveNeighbors(board, m, n, i, j);

            // In the beginning, every 2nd bit is 0;
            // So we only need to care about when will the 2nd bit become 1.
            if (board[i][j] == 1 && lives >= 2 && lives <= 3) {  
                board[i][j] = 3; // Make the 2nd bit 1: 01 ---> 11
            }
            if (board[i][j] == 0 && lives == 3) {
                board[i][j] = 2; // Make the 2nd bit 1: 00 ---> 10
            }
        }
    }

    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            board[i][j] >>= 1;  // Get the 2nd state.
        }
    }
}

public int liveNeighbors(int[][] board, int m, int n, int i, int j) {
    int lives = 0;
    for (int x = Math.max(i - 1, 0); x <= Math.min(i + 1, m - 1); x++) {
        for (int y = Math.max(j - 1, 0); y <= Math.min(j + 1, n - 1); y++) {
            lives += board[x][y] & 1;
        }
    }
    lives -= board[i][j] & 1;
    return lives;
}