C++ O(mn)-time, O(1)-space sol

  • 30

    First this solution does not involve bit-manipulation and only involves addition of integers.

    The idea is to go through the matrix from top-left corner to the bottom-right corner, and check only 4 cells ("accumulate" scores "for both cells" if the other cell is originally a 1). Graphically speaking, it is like this:

    O O O 
    O @ X
    X X X

    where the @ cell is the one that you are working on, 0 cells are those you have gone through (don't work on them again!), and the X cells are those you have not gone through and should work on. For example, if one X cell is originally a 1, you should add C (a constant) to @ cell, and simultaneously if @ cell is originally a 1, you add C to that X cell.

    The constant C can be 2. If it is 2 then when you will find that after done working on the current cell if it's value is 5 or 7 (cell @ is originally a 1 and have 2 or 3 neighbours) or 6 (cell @ is originally a 0 and have 3 neighbours), then you should reset it to be 1 (live), otherwise reset it to be zero (dead). And when accumulating scores, you know a cell is originally a 1 if it has odd-numbered score, and it is originally a 0 if it has even-numbered score. The code is

    class Solution {
        void gameOfLife(vector<vector<int>>& board) {
            if(board.empty()) return;
            const int m = board.size();
            const int n = board[0].size();
            for(int i=0; i<m; i++) {
                for(int j=0; j<n; j++) {
                    if(board[i][j]>=5 && board[i][j]<=7) board[i][j]=1;
                    else board[i][j]=0;
        void check(vector<vector<int>>& board, int i, int j, int a, int b) {
            const int m = board.size();
            const int n = board[0].size();
            if(a>=m || b<0 || b>=n) return;
            if(board[i][j]%2!=0) board[a][b]+=2;
            if(board[a][b]%2!=0) board[i][j]+=2;

  • 0

    Great idea with adding a Constant 2.

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