Game of Life 【leetcode】【98.9】【M】

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?

生命游戏中，对于任意细胞，规则如下：

每个细胞有两种状态-存活或死亡，每个细胞与以自身为中心的周围八格细胞产生互动。（如图，黑色为存活，白色为死亡）

当前细胞为存活状态时，当周围低于2个（不包含2个）存活细胞时， 该细胞变成死亡状态。（模拟生命数量稀少）
当前细胞为存活状态时，当周围有2个或3个存活细胞时， 该细胞保持原样。
当前细胞为存活状态时，当周围有3个以上的存活细胞时，该细胞变成死亡状态。（模拟生命数量过多）
当前细胞为死亡状态时，当周围有3个存活细胞时，该细胞变成存活状态。 （模拟繁殖）

可以把最初的细胞结构定义为种子，当所有在种子中的细胞同时被以上规则处理后, 可以得到第一代细胞图。按规则继续处理当前的细胞图，可以得到下一代的细胞图，周而复始。

自己采用的办法感觉好笨，还是在周围搞了一圈零，可以看看人家的：

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;

}

[思路]

inplace的话, 只要有办法区分 4种状态, DEAD->LIVE, DEAD->DEAD, LIVE->LIVE, LIVE->DEAD 即可. int 完全可以找4个数来表示这4种状态.

class Solution(object):
def cal(self,b,i,j):
#print b[i-1][j-1],b[i-1][j],b[i-1][j+1],',',b[i][j-1],b[i][j],b[i][j+1],',',b[i+1][j-1],b[i+1][j],b[i+1][j+1]
return b[i-1][j-1]+b[i-1][j]+b[i-1][j+1]+b[i][j-1]+b[i][j+1]+b[i+1][j-1]+b[i+1][j]+b[i+1][j+1]

def gameOfLife(self, board):
#print board

b = board

m = len(b)
if m == 0:
return
n = len(b[0])

b = [[0]*(n+2)]#list()#[[]]#[[0]]*n
#print b
for i in range(0,m):
#print board[i][:]
b.append([0])
b[i+1][:] += (board[i][:])
b[i+1] += [0]
# for j in board[i]:
#     b[i].append(j)
b.append([0]*(n+2))
#print b

for i in range(1,m+1):

for j in range(1,n+1):
cal = self.cal(b,i,j)
#print b[i][j],cal,' '
if b[i][j] == 1:
if cal < 2:
board[i-1][j-1] = 0
elif cal > 3:
board[i-1][j-1] = 0
else:
board[i-1][j-1] = b[i][j]
else:
if cal == 3:
board[i-1][j-1] = 1
else:
board[i-1][j-1] = b[i][j]

#print ''

#print board
"""
:type board: List[List[int]]
:rtype: void Do not return anything, modify board in-place instead.
"""