LeetCode – Game of Life (Java)
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:
Any live cell with fewer than two live neighbors dies, as if caused by underpopulation.
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 overpopulation..
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.
Java Solution 1
Because we need to solve the problem in place, we can use the higher bit to record the next state. And at the end, shift right a bit to get the next state for each cell.
public void gameOfLife(int[][] board) { if(board==null  board.length==0board[0].length==0) return; int m=board.length; int n=board[0].length; int[] x = {1, 1, 0, 1, 1, 1, 0, 1}; int[] y = {0, 1, 1, 1, 0, 1, 1, 1}; for(int i=0; i<m; i++){ for(int j=0; j<n; j++){ int count=0; for(int k=0; k<8; k++){ int nx=i+x[k]; int ny=j+y[k]; if(nx>=0&&nx<m&&ny>=0&&ny<n&&(board[nx][ny]&1)==1){ count++; } } //<2 die if(count<2){ board[i][j] &= 1; } //same state if(count==2count==3){ board[i][j] = board[i][j]<<1; } //go live if(count==3){ board[i][j] =2; } //>3 die if(count>3){ board[i][j] &=1; } } } for(int i=0; i<m; i++){ for(int j=0; j<n; j++){ board[i][j] = board[i][j]>>1; } } } 
<pre><code> String foo = "bar"; </code></pre>

Mewanbanjop Mawroh