题目描述

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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

分析

参考：leetcode 289: Game of Life（西施豆腐渣博客）

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。

下面的图片，是一个Game of Life变化图。

代码

    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;
    }