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// Source : https://leetcode.com/problems/game-of-life/
// Author : cornprincess
// Date : 2020-04-02
/*****************************************************************************************************
*
* 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.
* The next state is created by applying the above rules simultaneously to every cell in the current
* state, where births and deaths occur simultaneously.
*
* Example:
*
* Input:
* [
* [0,1,0],
* [0,0,1],
* [1,1,1],
* [0,0,0]
* ]
* Output:
* [
* [0,0,0],
* [1,0,1],
* [0,1,1],
* [0,1,0]
* ]
*
* 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?
*
******************************************************************************************************/
package GameOfLife;
public class GameOfLife {
// Time Complexity O(nm)
// Space Complexity O(mn)
public void copyBoard(int[][] board) {
int[] dx = {-1, -1, -1, 0, 0, 1, 1, 1};
int[] dy = {-1, 0, 1, -1, 1, -1, 0, 1};
int rows = board.length;
int cols = board[0].length;
int[][] copyBoard = new int[rows][cols];
for (int row = 0; row < rows; row++) {
for (int col = 0; col < cols; col++) {
copyBoard[row][col] = board[row][col];
}
}
for (int row = 0; row < rows; row++) {
for (int col = 0; col < cols; col++) {
int liveNeighbour = 0;
for (int i = 0; i < 8; i++) {
int x = row + dx[i];
int y = col + dy[i];
if (x >= rows || x < 0 || y >= cols || y < 0) {
continue;
}
liveNeighbour += copyBoard[x][y] & 1;
}
if (board[row][col] == 1) {
if (liveNeighbour < 2 || liveNeighbour > 3) {
board[row][col] = 0;
}
} else {
if (liveNeighbour == 3) {
board[row][col] = 1;
}
}
}
}
}
// Time Complexity O(nm)
// Space Complexity O(1)
public void bitwise(int[][] board) {
int[] dx = {-1, -1, -1, 0, 0, 1, 1, 1};
int[] dy = {-1, 0, 1, -1, 1, -1, 0, 1};
int rows = board.length;
int cols = board[0].length;
for (int row = 0; row < rows; row++) {
for (int col = 0; col < cols; col++) {
int liveNeighbour = 0;
for (int i = 0; i < 8; i++) {
int x = row + dx[i];
int y = col + dy[i];
if (x >= rows || x < 0 || y >= cols || y < 0) {
continue;
}
liveNeighbour += board[x][y] & 1;
}
if (board[row][col] == 1) {
if (liveNeighbour == 2 || liveNeighbour == 3) {
board[row][col] |= 2;
}
} else {
if (liveNeighbour == 3) {
board[row][col] |= 2;
}
}
}
}
for (int row = 0; row < rows; row++) {
for (int col = 0; col < cols; col++) {
board[row][col] >>= 1;
}
}
}
}