Medium
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?
1ms:
public void gameOfLife(int[][] board) {
final int L2D = 1;
final int L2L = 3;
final int D2L = 2;
final int D2D = 0;
int m = board.length;
int n = board[0].length;
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
// System.out.println("i="+i+",j="+j);
int count = 0;
int status = board[i][j]&1;
// System.out.println("status="+status);
count+=getR(i,j,m,'m',board)&1;
count+=getR(i,j,m,'p',board)&1;
count+=getC(i,j,n,'m',board)&1;
count+=getC(i,j,n,'p',board)&1;
count+=getD(i-1,j,m,n,'m',board)&1;
count+=getD(i-1,j,m,n,'p',board)&1;
count+=getD(i+1,j,m,n,'m',board)&1;
count+=getD(i+1,j,m,n,'p',board)&1;
// System.out.println("count="+count);
if(status==1){
if(count<4&&count>1)
board[i][j]=L2L;
else
board[i][j]=L2D;
}else{
if(count==3)
board[i][j]=D2L;
else
board[i][j]=D2D;
}
}
}
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
board[i][j]=board[i][j]>>1;
}
}
}
int getC(int i,int j,int max, char x,int[][] board){
if(x=='m'){
if(j==0)
return 0;
else
return board[i][j-1];
}else{
if(j==max-1)
return 0;
else
return board[i][j+1];
}
}
int getR(int i,int j,int max,char x,int[][] board){
if(x=='m'){
if(i==0)
return 0;
else
return board[i-1][j];
}else{
if(i==max-1)
return 0;
else
return board[i+1][j];
}
}
int getD(int i,int j,int m,int n,char x,int[][] board){
if(x=='m'){
if(i<0||j==0||i==m)
return 0;
else
return board[i][j-1];
}else{
if(i<0||j==n-1||i==m)
return 0;
else
return board[i][j+1];
}
}
1ms:
public void gameOfLife(int[][] board){
int m = board.length;
int n = m>0?board[0].length:0;
if(m*n==0) return;
// System.out.println(m+":"+n);
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
// System.out.println("i:"+i+",j:"+j);
int lives = getLives(i,j,m,n,board);
if(board[i][j]==1&&lives>1&&lives<4)
board[i][j]=3;
if(board[i][j]==0&&lives==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;
}
}
}
int getLives(int i,int j,int m, int n,int[][] board){
int lives = 0;
for(int a=Math.max(i-1, 0);a<=Math.min(i+1, m-1);a++){
for(int b=Math.max(j-1, 0);b<=Math.min(j+1, n-1);b++){
// System.out.println("a:"+a+",b:"+b);
lives+=board[a][b]&1;
}
}
lives-=board[i][j]&1;
return lives;
}
本文介绍了一种基于康威生命游戏规则的算法实现方法。该算法通过计算每个细胞周围活细胞的数量来决定其生死状态,并据此更新整个游戏板的状态。文章提供了两种不同的Java实现方案,包括状态位操作和邻居计数的方法。
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