Leetcode 题目解析之 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.

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?

题目中说最好是in-place，因为board是int型的，所以要区分四种状态（dead是0，live是1）：

dead->live 01

dead->dead 00

live->dead 10

live->live 11

或者直接用0,1,2,3四个数字表示四种状态都可以，然后通过移位进行结果的更新。

    public void gameOfLife(int[][] board) {
        if (board == null || board.length == 0 || board[0] == null
                || board[0].length == 0)
            return;
        int m = board.length;
        int n = board[0].length;
        // 判断每个点，下一时刻的状态
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int x = getLiveNum(board, i, j);
                if (board[i][j] == 0) {
                    if (x == 3)
                        board[i][j] += 10;
                } else {
                    if (x == 2 || x == 3)
                        board[i][j] += 10;
                }
            }
        }
        // 更新每个点的状态
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                board[i][j] /= 10;
            }
        }
    }
    // 获取board[x][y]邻居live的数量
    private int getLiveNum(int[][] board, int x, int y) {
        int c = 0;
        for (int i = x - 1; i <= x + 1; i++) {
            for (int j = y - 1; j <= y + 1; j++) {
                if (i < 0 || j < 0 || i > board.length - 1
                        || j > board[0].length - 1 || (i == x && j == y))
                    continue;
                if (board[i][j] % 10 == 1)
                    c++;
            }
        }
        return c;
    }