289. Game of Life

According to the 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."

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 (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.

Write a function to compute the next state (after one update) of the board given its current state. The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously.

Example:

Input: 
[
  [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]
]

Follow up:

1. Could you solve it in-place? Remember that the board needs to be updated at the same time: You cannot update some cells first and then use their updated values to update other cells.

2. 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 the border of the array. How would you address these problems?

Brute force解法可以使用一個暫時的2D array去存新的一輪的狀態，然後掃瞄完當前這個格子後再把暫時的2D array存到input 2D array裡，但因為題目要求in-place解，所以採用bit manipulation來記錄新舊的狀態。右邊的bit存舊的狀態，左邊的bit存新的狀態。解法中count變數在算的時候會把當前的格子本身是0還是1也算進去，如果是1的話，count會多1，因為我們目標只要看當前格子周遭8個格子。

void gameOfLife(vector<vector<int>>& board) { // time: O(m*n); space: O(1)
    int m = board.size(), n = board[0].size();
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            int count = 0;
            for (int I = max(i - 1, 0); I < min(i + 2, m); ++I) {
                for (int J = max(j - 1, 0); J < min(j + 2, n); ++J) {
                    // count includes the current cell itself
                    count += board[I][J] & 1;
                }
            }
            // 00 -> 10
            // 01 -> 11
            if (count == 3 || count - board[i][j] == 3) board[i][j] |= 2;
        }
    }
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            board[i][j] >>= 1;
        }
    }
}