中山网_六盘水网络推广_西安百度_网络推广的方法

题目描述

题解思路

我们记录上一行和当前行转换之后的状态，当前行转换之后的状态计算完毕后调整上一行状态，直至最后一行状态计算完毕后调整最后一行状态

题解代码

pub fn game_of_life(board: &mut Vec<Vec<i32>>) {let (m, n) = (board.len(), board[0].len());let mut rows = vec![vec![0; n].clone(); 2];let live = |board: &Vec<Vec<i32>>, i: usize,  j: usize| -> i32 {let mut lives = 0;// 上if i > 0 {lives += board[i - 1][j];}// 下if i + 1 < m {lives += board[i + 1][j];}// 左if j > 0 {lives += board[i][j - 1];// 左上if i > 0 {lives += board[i - 1][j - 1];}// 左下if i + 1 < m {lives += board[i + 1][j - 1];}}// 右if j + 1 < n {lives += board[i][j + 1];// 右上if i > 0 {lives += board[i - 1][j + 1];}// 右下if i + 1 < m {lives += board[i + 1][j + 1];}}if board[i][j] == 1 {if lives < 2 || lives > 3 {0} else {1}} else {if lives == 3 {1} else {0}}};for i in 0..n {rows[0][i] = live(board, 0, i);}for i in 1..m {for j in 0..n {rows[i % 2][j] = live(board, i, j);}for j in 0..n {board[i - 1][j] = rows[(i - 1) % 2][j];}}for i in 0..n {board[m - 1][i] = rows[(m - 1) % 2][i];}
}

题目链接

https://leetcode.cn/problems/game-of-life/description/