标签：
https://leetcode.com/problems/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):
Write a function to compute the next state (after one update) of the board given its current state.
Follow up:
struct neighbor{
int dies, lives;
neighbor() {dies = 0; lives = 0;}
neighbor(int d, int l): dies(d), lives(l) {}
};
class Solution {
public:
bool check(int m, int n, int a, int b) {
if(a<0 || a>=m || b<0 || b>=n) return false;
return true;
}
void gameOfLife(vector<vector<int>>& board) {
if(board.size() == 0) return;
int m = board.size(), n = board[0].size(), ni, nj;
int dir[8][2] = {{-1,0},{-1,-1},{-1,1},{0,-1},{0,1},{1,0},{1,-1},{1,1}};
vector<vector<int> > res(m, vector<int>(n));
neighbor neg[m][n];
for(int i=0;i<m;++i) {
for(int j=0;j<n;++j) {
for(int k=0;k<8;++k) {
ni = i + dir[k][0]; nj = j + dir[k][1];
if(check(m, n, ni, nj)) {
if(board[ni][nj] == 1) ++neg[i][j].lives;
else ++neg[i][j].dies;
}
}
}
}
for(int i=0;i<m;++i) {
for(int j=0;j<n;++j) {
if(board[i][j] && neg[i][j].lives < 2) res[i][j] = 0;
else if(board[i][j] && (neg[i][j].lives == 2 || neg[i][j].lives == 3)) res[i][j] = 1;
else if(board[i][j] && neg[i][j].lives > 3) res[i][j] = 0;
else if(!board[i][j] && neg[i][j].lives == 3) res[i][j] = 1;
else res[i][j] = board[i][j];
}
}
board = res;
}
};
标签：
原文地址：http://www.cnblogs.com/fu11211129/p/5018138.html
