Leetcode:game of Life-Conway games

1. Title

Game of Life (Conway Lifetime game)

2. Address of the topic

Https://leetcode.com/problems/game-of-life

3. Topic content

English:

According to the Wikipedia's article: "The Gameof Life, also known simply as Life , is a cellular automa ton devised by the British mathematician John Horton Conway in 1970. "

Given a board with m - n cells, each cell have 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 t He above Wikipedia article):

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

2. Any live cell with a 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 cells 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 it current state.

Chinese:

According to Wikipedia article Conway's Game of Life (Conway Lifetime game), Conway Life Game is a cellular automaton invented by British mathematician John Hoton Kangwei in 1970.

A m*n cell matrix is given, with each cell having an initial state: Survival (1) or death (0). Each cell changes to 8 cells around it, and the rules are as follows:

1. When the current cell is alive, the cell becomes dead when there are less than 2 surviving cells in the surrounding state. (Simulation of the number of life is scarce)

2. When the current cell is alive, the cell remains intact when there are 2 or 3 surviving cells around it.

3. When the current cell is alive, the cell becomes dead when there are more than 3 surviving cells in the surrounding state. (excessive number of simulated life)

4. When the current cell is in the dead state, the cell becomes viable when there are exactly 3 surviving cells around it. (Simulated reproduction)

Write a function that calculates the next state of the cell matrix based on the current state of the Matrix.

4. Methods of Solving problems

Without using the definition of a new matrix, in order to solve this problem, we need to define a set of intermediate states, the intermediate state requirements can see a cell change before and after two aspects of life and death situation. The Java code is as follows:

/** * @ function Description: leetcode 289 - game of life * @ developer:tsybius2014  * @ Development Time: October 8, 2015  */public class solution {    //death Unit      final int dead = 0;    //Survival Unit      final int alive = 1;    //change: death → Death      final int dead_to_dead = 0;    //change: Survival → Survival      final int alive_to_alive = 1;    //change: Survival → Death      final int alive_to_dead = 2;    //change: death → Survival      final int dead_to_alive = 3;        /**      *  determine if a point is dead before the change in this round      *  @param  obj      *  @return &Nbsp;    */    private boolean isaliveold (Int obj)  {        if  (obj == alive_to_alive | |  obj == alive_to_dead)  {             return true;        }         else {            return  false;        }    }    /**      *  determine if a point is dead after the change in this round      *  @param  obj      *  @return      */    private  boolean isalivenew (int obj)  {        if  (obj &NBSP;%&NBSP;2&NBSP;==&NBSP;1)  {            return true;         } else {             return false;        }    }         /**     *  Life Games       *  @param  board     */    public void  Gameoflife (int[][] board)  {        //input Legality check          if  (board == null)  {             return;        }         int height = board.length;         if  (Height == 0)  {            return;         }        int  width = board[0].length;        if  (width ==  0)  {            return;         }        //examines all points and changes their life state          int counter = 0;         for  (int i = 0; i < height; i++)  {             for  (int j = 0; j <  width; j++)  {                                 //statistics surrounding life conditions                  counter = 0;                 if  (i >  0 && j > 0 && isaliveold (board[i - 1][j  -&NBSP;1])  {                     counter++;                 }                 if  (I > 0 && isaliveold (BOARD[I&NBSP;-&NBSP;1][J))  {                     counter++;                }                 if  (i >  0 && j < width - 1 && isaliveold (board[i - &NBSP;1][J&NBSP;+&NBSP;1])  {                     counter++;                 }                 if  (J > 0 && isaliveold (board[i][j &NBSP;-&NBSP;1])  {                     counter++;                 }                 if  (J  < width - 1 && isaliveold (board[i][j + 1])  {                      counter++;                 }                if   (i < height - 1 && j > 0 &&  Isaliveold (board[i + 1][j - 1])  {                     counter++;                 }                 if  (I < height - 1 && isaliveold (board[i &NBSP;+&NBSP;1][J])  {                     counter++;                 }                 if  (i < height - 1 && j <  width - 1 && isaliveold (board[i + 1][j + 1])  {                      counter++;                 }                                 //determines the current point change based on life conditions around the specified point                  if  (Isaliveold (Board[i][j]) )  {                     if  (counter < 2)  {                         //1. Survival units around the survival unit of less than 2, the unit died                          board[i][j] = ALIVE_TO_DEAD;                     }  else if  (counter == 2 | | &NBSP;COUNTER&NBSP;==&NBSP;3)  {                         //2. There are 2-3 surviving units around the surviving unit, and the unit continues to survive                          board[i][j] = ALIVE_TO_ALIVE;                      } else {               &NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;//3. Survival units around the surviving Unit 3 extra, the unit died                           board[i][j] = ALIVE_TO_DEAD;                     }                 } else {                     if  (counter == 3)  {                         &NBSP;//4. The surviving unit around the death unit is exactly 3, and the unit becomes a viable State                          board[i][j] = DEAD_TO_ALIVE;                                           } else {                         board[i][j] = dead_to_ Dead;                    }                 }             }        }         //the life and death of each point according to the transformed survival status          for  (int i = 0; i < height; i++)  {             for  (int j = 0; j <  width; j++)  {                 if  (Isalivenew (board[i][j))  {                     board[i][j] = ALIVE;                 } else {                     board[i][j] =  dead;                }             }         }    }}

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Leetcode:game of Life-Conway games