Game of Life

https://leetcode.com/problems/game-of-life/description/

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):

Any live cell with fewer than two live neighbors dies, as if caused by under-population.

Any live cell with two or three live neighbors lives on to the next generation.

Any live cell with more than three live neighbors dies, as if by over-population..

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:

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.

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?

Thoughts

一道细胞状态机问题. 要求每个细胞根据它们的邻居的状态更新自己的状态, 并且要求能够同时更新, 不能先更新一些再更新一些 (会导致状态混乱). 因此当更新一个细胞状态时, 要知道周围细胞的原状态. 但周围细胞可能已经被更新过了, 还怎么知道它们的原状态呢? 解决方案是除了0和1外再多维持两个状态, 2代表从1要变到0, 3代表要从0变到1.

Code

class Solution {
    public void gameOfLife(int[][] board) {
        int[][] dirs = new int[][]{{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
        int m = board.length, n = board[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int lives = 0;
                for (int[] dir : dirs) {
                    int x = i + dir[0];
                    int y = j + dir[1];
                    if (x >= 0 && x < m && y >= 0 && y < n) {
                        if (board[x][y] == 1 || board[x][y] == 2) {
                            lives++; 
                        }
                    }
                }

                if (board[i][j] == 1 && (lives < 2 || lives > 3)) {
                    board[i][j] = 2;
                } else if (board[i][j] == 0 && lives == 3) {
                    board[i][j] = 3;
                }
            }
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] % 2 == 0) {
                    board[i][j] = 0;
                } else {
                    board[i][j] = 1;
                }
            }
        }
    }
}

Analysis

时间复杂度O(MN).