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GameOfLife289.java
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GameOfLife289.java
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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):
*
* 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?
*/
public class GameOfLife289 {
public void gameOfLife(int[][] board) {
if (board == null || board.length == 0 || board[0].length == 0) return;
int n = board.length;
int m = board[0].length;
int[][] newBoard = new int[n][m];
for (int i=0; i<n; i++) {
for (int j=0; j<m; j++) {
if (board[i][j] == 0) {
newBoard[i][j] = countLive(board, i, j) == 3 ? 1 : 0;
} else {
int lives = countLive(board, i, j);
newBoard[i][j] = (lives == 2 || lives == 3) ? 1 : 0;
}
}
}
for (int i=0; i<n; i++) {
for (int j=0; j<m; j++) {
board[i][j] = newBoard[i][j];
}
}
}
private int countLive(int[][] board, int x, int y) {
int n = board.length;
int m = board[0].length;
int res = 0;
for (int i=x-1; i<=x+1; i++) {
if (i < 0 || i >= n) continue;
for (int j=y-1; j<=y+1; j++) {
if (j < 0 || j >= m || (i == x && j == y)) continue;
if (board[i][j] == 1) res++;
}
}
return res;
}
/**
* https://leetcode.com/problems/game-of-life/discuss/73223/Easiest-JAVA-solution-with-explanation
*/
public void gameOfLife2(int[][] board) {
if (board == null || board.length == 0) return;
int m = board.length, n = board[0].length;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int lives = liveNeighbors(board, m, n, i, j);
// In the beginning, every 2nd bit is 0;
// So we only need to care about when will the 2nd bit become 1.
if (board[i][j] == 1 && lives >= 2 && lives <= 3) {
board[i][j] = 3; // Make the 2nd bit 1: 01 ---> 11
}
if (board[i][j] == 0 && lives == 3) {
board[i][j] = 2; // Make the 2nd bit 1: 00 ---> 10
}
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
board[i][j] >>= 1; // Get the 2nd state.
}
}
}
public int liveNeighbors(int[][] board, int m, int n, int i, int j) {
int lives = 0;
for (int x = Math.max(i - 1, 0); x <= Math.min(i + 1, m - 1); x++) {
for (int y = Math.max(j - 1, 0); y <= Math.min(j + 1, n - 1); y++) {
lives += board[x][y] & 1;
}
}
lives -= board[i][j] & 1;
return lives;
}
public void gameOfLife3(int[][] board) {
int M = board.length;
int N = board[0].length;
for (int i=0; i<M; i++) {
for (int j=0; j<N; j++) {
update(board, i, j, M, N);
}
}
for (int i=0; i<M; i++) {
for (int j=0; j<N; j++) {
refresh(board, i, j);
}
}
}
private void update(int[][] board, int i, int j, int M, int N) {
int lives = 0;
for (int x=i-1; x<=i+1; x++) {
if (x < 0 || x >= M) continue;
for (int y=j-1; y<=j+1; y++) {
if (y < 0 || y >= N || (x == i && y == j)) continue;
if ((board[x][y] & 1) == 1) lives++;
}
}
boolean wasLive = (board[i][j] & 1) == 1;
if (wasLive) {
if (lives == 2 || lives == 3) {
board[i][j] |= 1 << 1;
}
} else {
if (lives == 3) {
board[i][j] |= 1 << 1;
}
}
}
private void refresh(int[][] board, int i, int j) {
board[i][j] = board[i][j] >> 1;
}
}