Leetcode 289 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?

思路：要求 in-place ，所以要区分三种变化状态 1->0,1->1,0->1，将三种状态分别用-1/-2/-3三个状态值来表示

　　第一次遍历时，先判断当前cell的转换状态，且将cell值置为对应状态值，在后面的遍历中根据状态值则可推断出原始值，进而判断转换状态。

　　第二次遍历，则将状态值置为变换后的值，注意一下在每个cell遍历其周围八个点时用的循环，是怎么排除边界情况的

public class S289 {
    public void gameOfLife(int[][] board) {
        int m = board.length,n = board[0].length;
        for (int i = 0;i < m;i++) {
            for (int j = 0;j < n;j++) {
                int liveCount = 0;
                for (int k = i-1;k <= i+1;k++) {
                    for (int l = j-1;l <= j+1;l++) {
                        if(k < 0 || l < 0 || k > m-1 || l > n-1 || (k == i && l == j)) continue;
                        liveCount += getRealNum(board[k][l]);
                    }
                }
                if (board[i][j] == 1) {
                    if (liveCount <2 || liveCount >3 ) {
                        board[i][j] = -1;
                    } else {
                        board[i][j] = -2;
                    }
                } else {
                    if (liveCount == 3) {
                        board[i][j] = -3;
                    }
                }
            }
        }
        for (int i = 0;i < m;i++) {
            for (int j = 0;j < n;j++) {
                if (board[i][j] == -2 || board[i][j] == -3)
                    board[i][j] = 1;
                else if (board[i][j] == -1)
                    board[i][j] = 0;
            }
        }
    }
    public static int getRealNum(int i){
        if(i == -1 || i == -2)
            return 1;
        else if (i == -3) {
            return 0;
        }
        return i;
    }
    public static void main(String[] args) {
        S289 s = new S289();
        int [][]b = {{1}};
        s.gameOfLife(b);
    }
}