[Leetcode] 289. Game of Life 解题报告

题目：

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.

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?

思路：

没有特别的算法技巧，我在实现中觉得需要注意如下几个方面：1）为了做到in-place，可以对board中的数进行按位更新：也就是说下一个状态的数我们存储在board[i][j]的次低位，这样就可以保证不和当前状态发生冲突。2）获取邻居中live的个数的时候，需要注意边界情况并且把自身的live情况排除在外。

代码：

class Solution {public:    void gameOfLife(vector<vector<int>>& board) {        if (board.size() == 0 || board[0].size() == 0) {            return;        }        int row_num = board.size(), col_num = board[0].size();        for (int i = 0; i < row_num; ++i) {            for (int j = 0; j < col_num; ++j) {                int live_num = getLiveNeighborNumber(board, i, j);                if (board[i][j] % 2 == 1) {     // live                    if (live_num >= 2 && live_num <= 3) {                        board[i][j] += 2;                    }                }                else {                    if (live_num == 3) {                        board[i][j] += 2;                    }                }            }        }        for (int i = 0; i < row_num; ++i) {            for (int j = 0; j < col_num; ++j) {                board[i][j] >>= 1;            }        }    }private:    int getLiveNeighborNumber(vector<vector<int>> &board, int row, int col) {        int ret = 0, row_num = board.size(), col_num = board[0].size();        for (int i = -1; i <= 1; ++i) {            for (int j = -1; j <= 1; ++j) {                if (i == 0 && j == 0) {                    continue;                }                int y = row + i, x = col + j;                if (x < 0 || x >= col_num || y < 0 || y >= row_num) {                    continue;                }                ret += (board[y][x] % 2);            }        }        return ret;    }};