Game of Life – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
According to 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.”
The board is made up of an m x n grid of cells, where each cell has an initial state: live (represented by a 1) or dead (represented by a 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.
The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously. Given the current state of the m x n grid board, return the next state.
Example 1:
Input: board = [[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]]
Example 2:
Input: board = [[1,1],[1,0]] Output: [[1,1],[1,1]]
Constraints:
m == board.lengthn == board[i].length1 <= m, n <= 25board[i][j]is0or1.
C++ Game of Life LeetCode Solution
void gameOfLife(vector<vector<int>>& board) {
int m = board.size(), n = m ? board[0].size() : 0;
for (int i=0; i<m; ++i) {
for (int j=0; j<n; ++j) {
int count = 0;
for (int I=max(i-1, 0); I<min(i+2, m); ++I)
for (int J=max(j-1, 0); J<min(j+2, n); ++J)
count += board[I][J] & 1;
if (count == 3 || count - board[i][j] == 3)
board[i][j] |= 2;
}
}
for (int i=0; i<m; ++i)
for (int j=0; j<n; ++j)
board[i][j] >>= 1;
}
Java Game of Life LeetCode Solution
public void gameOfLife(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;
}
Python 3 Game of Life LeetCode Solution
def gameOfLife(self, board):
m,n = len(board), len(board[0])
for i in range(m):
for j in range(n):
if board[i][j] == 0 or board[i][j] == 2:
if self.nnb(board,i,j) == 3:
board[i][j] = 2
else:
if self.nnb(board,i,j) < 2 or self.nnb(board,i,j) >3:
board[i][j] = 3
for i in range(m):
for j in range(n):
if board[i][j] == 2: board[i][j] = 1
if board[i][j] == 3: board[i][j] = 0
def nnb(self, board, i, j):
m,n = len(board), len(board[0])
count = 0
if i-1 >= 0 and j-1 >= 0: count += board[i-1][j-1]%2
if i-1 >= 0: count += board[i-1][j]%2
if i-1 >= 0 and j+1 < n: count += board[i-1][j+1]%2
if j-1 >= 0: count += board[i][j-1]%2
if j+1 < n: count += board[i][j+1]%2
if i+1 < m and j-1 >= 0: count += board[i+1][j-1]%2
if i+1 < m: count += board[i+1][j]%2
if i+1 < m and j+1 < n: count += board[i+1][j+1]%2
return count
Array-1180
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