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 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.

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==0||board[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==2||count==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;
 
        }
    }
}

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Category >> Algorithms  
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  • coddee


    public class Solution {

    public void gameOfLife(int[][] b) {

    if (b == null || b.length == 0) {
    return;
    }

    int[] dx = new int[] {
    0,
    -1,
    -1,
    -1,
    0,
    1,
    1,
    1
    };
    int[] dy = new int[] {
    -1, -1, 0, 1, 1, 1, 0, -1
    };
    int m = b.length;
    int n = b[0].length;

    for (int i = 0; i < m; i++) {

    for (int j = 0; j < n; j++) {

    int count = 0;
    for (int k = 0; k = 0 && x = 0 && y < n && (b[x][y] & 1) == 1) {
    count++;
    }
    }
    if ((b[i][j] & 1) == 1) {
    if (count == 2 || count == 3) {
    b[i][j] |= 1 << 1;
    }
    } else {
    if (count == 3) {
    b[i][j] |= 1 << 1;
    }
    }

    }
    }

    for (int i = 0; i < m; i++) {
    for (int j = 0; j > 1;

    }
    }

    }
    }

  • Mewanbanjop Mawroh