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L289.java
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L289.java
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package Algorithm;
/*
* 因为此题目需要改变二维数组里面的值,所以会影响到对周围存活节点的计算。一定需要下面这样编码
* 0 : 上一轮是0,这一轮过后还是0
1 : 上一轮是1,这一轮过后还是1--本来上一轮是0,下轮是1表示为1,但是这样是为了保持原样的时候写代码简单
2 : 上一轮是1,这一轮过后变为0
3 : 上一轮是0,这一轮过后变为1
因此若是0,1则之前是死亡状态;2,3则之前是存活状态
*/
public class L289 {
public void gameOfLife(int[][] board) {
if(board[0].length == 0 || board == null) {
return ;
}
int row = board.length;
int col = board[0].length;
for(int i = 0; i < row; i ++) {
for(int j = 0; j < col; j ++) {
int count = neighborsBoard(board, row, col);
if(count == 2)
continue;
if(count == 3) {
board[i][j] = board[i][j] == 0 ? 3 : 1;
}else {
board[i][j] = board[i][j] == 1 ? 2 : 0;
}
}
}
//因为看变完后的,1有1 3 两种状态,0有0 2两种状态,经过模2运算后就可以知道是1还是0
for(int i = 0 ; i < row; i ++) {
for(int j = 0; j < col; j ++) {
board[i][j] = board[i][j] % 2;
}
}
}
public int neighborsBoard(int [][] board, int row, int col) {
int count = 0;
for(int i = row - 1; i < row + 1; i ++) {
for(int j = col - 1; j < col + 1; j ++) {
if(i == row && j == col) {
continue;
}
//这里board[i][j] == 1或者为2的原因是board[i][j]原来为1
if(i >= 0 && i < board.length && j >= 0 && j < board[0].length && (board[i][j] == 1 || board[i][j] == 2)) {
count ++;
}
}
}
return count;
}
}