package leetcode_1To300;

/**
 * 本代码来自 Cspiration，由 @Cspiration 提供
 * 题目来源：http://leetcode.com
 * - Cspiration 致力于在 CS 领域内帮助中国人找到工作，让更多海外国人受益
 * - 现有课程：Leetcode Java 版本视频讲解（1-900题）（上）（中）（下）三部
 * - 算法基础知识（上）（下）两部；题型技巧讲解（上）（下）两部
 * - 节省刷题时间，效率提高2-3倍，初学者轻松一天10题，入门者轻松一天20题
 * - 讲师：Edward Shi
 * - 官方网站：https://cspiration.com
 * - 版权所有，转发请注明出处
 */
public class _289_GameofLife {

    /**
     * 289. Game of Life
     * 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.

     time : O(m * n)
     space : O(1)

     * @param board
     */

    public void gameOfLife(int[][] board) {
        if (board == null || board.length == 0) return;
        int m = board.length;
        int n = board[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int count = countNeighbor(board, i, j);
                if (board[i][j] == 1) {
                    if (count == 2 || count == 3) {
                        board[i][j] += 2;
                    }
                } else if (count == 3) {
                    board[i][j] += 2;
                }
            }
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                board[i][j] = board[i][j] >> 1;
            }
        }
    }

    private int countNeighbor(int[][] board, int i, int j) {
        int count = 0;
        for (int row = Math.max(0, i - 1); row <= Math.min(i + 1, board.length - 1); row++) {
            for (int col = Math.max(0, j - 1); col <= Math.min(j + 1, board[0].length - 1); col++) {
                if (row == i && col == j) continue;
                if ((board[row][col] & 1) == 1) count++;
            }
        }
        return count;
    }

}