# Java 1ms O(mn) time(m*n is the size of the given array) naive and simple in-place solution

• The key point is how to make the updated value does not interfere with the current judgement.
In this problem, there are only 0 and 1 in each cell. Once certain condition is satisfied, 0 will be 1 or 1 will be 0. Otherwise, they keep their value. My solution uses -1 and -2 to mark the 0 and 1 which will not be changed after update and then rewrite the array. Since -1 is the marked 1, during counting 1s in the neighborhood, both 1 and -1 should be counted.
For the rewrite, when hitting a -1, it should be recovered to 1 while -2 should be recovered to 0. And when hitting 0 or 1, 0 should be flipped into 1 while 1 should be flipped into 0.

``````public class Solution {
public void gameOfLife(int[][] board) {
int m = board.length;
int n = 0;
if(m>0) n = board[0].length;
for(int i = 0;i<m;i++){
for(int j = 0;j<n;j++){
int count_1 = 0; //used to count the 1s in the current neighborhood
if(i>0){
if(board[i-1][j]==1||board[i-1][j]==-1) count_1++;
if(j>0){
if(board[i-1][j-1]==1||board[i-1][j-1]==-1) count_1++;
}
if(j<n-1){
if(board[i-1][j+1]==1||board[i-1][j+1]==-1) count_1++;
}
}
if(i<m-1){
if(board[i+1][j]==1||board[i+1][j]==-1) count_1++;
if(j>0){
if(board[i+1][j-1]==1||board[i+1][j-1]==-1) count_1++;
}
if(j<n-1){
if(board[i+1][j+1]==1||board[i+1][j+1]==-1) count_1++;
}
}
if(j>0){
if(board[i][j-1]==1||board[i][j-1]==-1) count_1++;
}
if(j<n-1){
if(board[i][j+1]==1||board[i][j+1]==-1) count_1++;
}
//if current entry is 1, mark it as -1
if(count_1>=2&&count_1<=3&&board[i][j]==1) board[i][j] = -1;
//if current entry is 0, mark it as -2
if(count_1!=3&&board[i][j] == 0) board[i][j] = -2;
}
}
for(int i = 0;i<m;i++){
for(int j = 0;j<n;j++){
if(board[i][j]==-1) board[i][j] = 1;
else if(board[i][j]==-2) board[i][j] = 0;
else if(board[i][j]==1) board[i][j] = 0;
else board[i][j] = 1;
}
}
}
}``````

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