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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):
Write a function to compute the next state (after one update) of the board given its current state.
Follow up:
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
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思路：
细胞的转化情况有四种：
dead -> dead
live->
dead
dead-> live
live-> live
由于题目要求用in-place就地操作。因此在转化的过程中，即要体现出转化结果也要保留转化之前的状态。
由于转化前、后都是两种状态，因此可以用2个bit来表示：[bit2, bit1]
bit1：转化前状态；
bit2：转化后状态。
因此，上面的转化情况表示为：
00:dead <- dead
01:dead <- live
10:live <- dead
11:live <- live
由于bit2默认为0，因此只考虑10和11两种情况即可。
board[i][j] & 1：表示取bit1；
board[i][j] >> 1：表示取bit2。
java code:
public class Solution {
public void gameOfLife(int[][] board) {
if(board == null || board.length == 0) return;
int rows = board.length;
int cols = board[0].length;
for(int i=0;i<rows;i++) {
for(int j=0;j<cols;j++) {
int lives = liveInNeighbors(board, rows, cols, i, j);
// 只考虑第2位0->1和1->1的情况即可，因为第2位默认为0
if(board[i][j]==0 && lives==3)
board[i][j] = 2;
if(board[i][j]==1 && (lives==2 || lives==3))
board[i][j] = 3;
}
}
for(int i=0;i<rows;i++) {
for(int j=0;j<cols;j++) {
board[i][j] >>= 1; // 取高位
}
}
}
private int liveInNeighbors(int board[][], int rows, int cols, int row, int col) {
int lives = 0;
for(int i = Math.max(0, row-1);i <= Math.min(rows-1, row+1);i++) {
for(int j = Math.max(0, col-1);j <= Math.min(cols-1, col+1);j++) {
lives += board[i][j] & 1; // 取低位
}
}
lives -= board[row][col] & 1;
return lives;
}
}
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原文地址：http://blog.csdn.net/itismelzp/article/details/51627386
