LeetCode289. Game of Life

文章目录

一、题目

According to 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."

The board is made up of an m x n grid of cells, where each cell has an initial state: live (represented by a 1) or dead (represented by a 0). Each cell interacts with its eight neighbors (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article):

Any live cell with fewer than two live neighbors dies as if caused by under-population.

Any live cell with two or three live neighbors lives on to the next generation.

Any live cell with more than three live neighbors dies, as if by over-population.

Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.

The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously. Given the current state of the m x n grid board, return the next state.

Example 1:

Input: board = [[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]]

Example 2:

Input: board = [[1,1],[1,0]]

Output: [[1,1],[1,1]]

Constraints:

m == board.length

n == board[i].length

1 <= m, n <= 25

board[i][j] is 0 or 1.

Follow up:

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

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 upon the border of the array (i.e., live cells reach the border). How would you address these problems?

二、题解

cpp 复制代码
class Solution {
public:
    int dirs[8][2] = {-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};
    void gameOfLife(vector<vector<int>>& board) {
        int m = board.size(),n = board[0].size();
        for(int i = 0;i < m;i++){
            for(int j = 0;j < n;j++){
                int liveNum = 0;
                //遍历周围8个位置
                for(int k = 0;k < 8;k++){
                    int x = i + dirs[k][0];
                    int y = j + dirs[k][1];
                    if(x < 0 || x >= m || y < 0 || y >= n) continue;
                    if(board[x][y] == 1 || board[x][y] == -1) liveNum++;
                }
                //更新状态
                if(board[i][j] == 1 && (liveNum < 2 || liveNum > 3)) board[i][j] = -1;
                else if(board[i][j] == 0 && liveNum == 3) board[i][j] = 2;
            }
        }
        //更新board
        for(int i = 0;i < m;i++){
            for(int j = 0;j < n;j++){
                if(board[i][j] > 0) board[i][j] = 1;
                else board[i][j] = 0;
            }
        }
    }
};
相关推荐
九圣残炎16 分钟前
【从零开始的LeetCode-算法】1456. 定长子串中元音的最大数目
java·算法·leetcode
lulu_gh_yu21 分钟前
数据结构之排序补充
c语言·开发语言·数据结构·c++·学习·算法·排序算法
丫头,冲鸭!!!41 分钟前
B树(B-Tree)和B+树(B+ Tree)
笔记·算法
Re.不晚1 小时前
Java入门15——抽象类
java·开发语言·学习·算法·intellij-idea
ULTRA??1 小时前
C加加中的结构化绑定(解包,折叠展开)
开发语言·c++
凌云行者1 小时前
OpenGL入门005——使用Shader类管理着色器
c++·cmake·opengl
凌云行者2 小时前
OpenGL入门006——着色器在纹理混合中的应用
c++·cmake·opengl
为什么这亚子2 小时前
九、Go语言快速入门之map
运维·开发语言·后端·算法·云原生·golang·云计算
2 小时前
开源竞争-数据驱动成长-11/05-大专生的思考
人工智能·笔记·学习·算法·机器学习
~yY…s<#>2 小时前
【刷题17】最小栈、栈的压入弹出、逆波兰表达式
c语言·数据结构·c++·算法·leetcode