289. Game of Life

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):

1. Any live cell with fewer than two live neighbors dies as if caused by under-population.
2. Any live cell with two or three live neighbors lives on to the next generation.
3. Any live cell with more than three live neighbors dies, as if by over-population.
4. 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.length
• n == board[i].length
• 1 <= m, n <= 25
• board[i][j] is 0 or 1.

Follow up:

• Could you solve it in-place? Remember that the board needs to be updated simultaneously: 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 upon the border of the array (i.e., live cells reach the border). How would you address these problems?

From: LeetCode
Link: 289. Game of Life

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Solution:

Ideas：

I first create a function countLiveNeighbors to count the live neighbors of a given cell.

In the gameOfLife function, I create a copy of the board to maintain the original state of the board while applying the rules.

Then, for each cell in the board, I calculate the number of live neighbors and apply the four rules given in the problem statement.

Finally, I deallocate the memory reserved for the copied board to avoid memory leaks.

Code:

int countLiveNeighbors(int** board, int boardSize, int* boardColSize, int row, int col) {int liveNeighbors = 0;int rowDirections[8] = {-1, -1, -1, 0, 0, 1, 1, 1};int colDirections[8] = {-1, 0, 1, -1, 1, -1, 0, 1};for (int i = 0; i < 8; i++) {int newRow = row + rowDirections[i];int newCol = col + colDirections[i];if (newRow >= 0 && newRow < boardSize && newCol >= 0 && newCol < boardColSize[row]) {liveNeighbors += board[newRow][newCol];}}return liveNeighbors;
}void gameOfLife(int** board, int boardSize, int* boardColSize) {int** copyBoard = (int**) malloc(boardSize * sizeof(int*));for (int i = 0; i < boardSize; i++) {copyBoard[i] = (int*) malloc(boardColSize[i] * sizeof(int));for (int j = 0; j < boardColSize[i]; j++) {copyBoard[i][j] = board[i][j];}}for (int row = 0; row < boardSize; row++) {for (int col = 0; col < boardColSize[row]; col++) {int liveNeighbors = countLiveNeighbors(copyBoard, boardSize, boardColSize, row, col);if (copyBoard[row][col] == 1 && (liveNeighbors < 2 || liveNeighbors > 3)) {board[row][col] = 0;}else if (copyBoard[row][col] == 0 && liveNeighbors == 3) {board[row][col] = 1;}}}for (int i = 0; i < boardSize; i++) {free(copyBoard[i]);}free(copyBoard);
}