zoukankan      html  css  js  c++  java
  • 529. Minesweeper

    use std::collections::LinkedList;
    
    /**
    529. Minesweeper
    https://leetcode.com/problems/minesweeper/
    Let's play the minesweeper game (Wikipedia, online game)!
    You are given an m x n char matrix board representing the game board where:
    . 'M' represents an unrevealed mine,
    . 'E' represents an unrevealed empty square,
    . 'B' represents a revealed blank square that has no adjacent mines (i.e., above, below, left, right, and all 4 diagonals),
    . digit ('1' to '8') represents how many mines are adjacent to this revealed square, and
    . 'X' represents a revealed mine.
    You are also given an integer array click where click = [clickr, clickc] represents the next click position among all the unrevealed squares ('M' or 'E').
    Return the board after revealing this position according to the following rules:
    1. If a mine 'M' is revealed, then the game is over. You should change it to 'X'.
    2. If an empty square 'E' with no adjacent mines is revealed, then change it to a revealed blank 'B' and all of its adjacent unrevealed squares should be revealed recursively.
    3. If an empty square 'E' with at least one adjacent mine is revealed, then change it to a digit ('1' to '8') representing the number of adjacent mines.
    4. Return the board when no more squares will be revealed.
    
    Example 1:
    Input: board = [["E","E","E","E","E"],["E","E","M","E","E"],["E","E","E","E","E"],["E","E","E","E","E"]], click = [3,0]
    Output: [["B","1","E","1","B"],["B","1","M","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]]
    
    Constraints:
    1. m == board.length
    2. n == board[i].length
    3. 1 <= m, n <= 50
    4. board[i][j] is either 'M', 'E', 'B', or a digit from '1' to '8'.
    5. click.length == 2
    6. 0 <= clickr < m
    7. 0 <= clickc < n
    8. board[clickr][clickc] is either 'M' or 'E'.
    */
    
    pub struct Solution {}
    
    impl Solution {
        /*
        Solution: BFS, scan 8 directions for every node, Time:O(m*n), Space:O(m*n);
        */
        pub fn update_board(mut board: Vec<Vec<char>>, click: Vec<i32>) -> Vec<Vec<char>> {
            let (m, n) = (board.len() as i32, board[0].len() as i32);
            //dir is 8 rows and 2 columns
            let mut dir: [[i32; 2]; 8] = [[-1, 0], [1, 0], [0, -1], [0, 1], [-1, -1], [1, 1], [1, -1], [-1, 1]];
            let mut queue: LinkedList<(usize, usize)> = LinkedList::new();
            //check the x,y of click if a Mine first
            queue.push_back((click[0] as usize, click[1] as usize));
            while !queue.is_empty() {
                let mut cur = queue.pop_front().unwrap();
                if board[cur.0 as usize][cur.1 as usize] == 'M' {
                    board[cur.0 as usize][cur.1 as usize] = 'X'
                } else {
                    //check 8 direction of current position and count the number of Mine
                    let mut count = 0;
                    for d in &dir {
                        let x = cur.0 as i32 + d[0];
                        let y = cur.1 as i32 + d[1];
                        if x < 0 || y < 0 || x >= m || y >= n {
                            //check border case
                            continue;
                        }
                        if board[x as usize][y as usize] == 'M' {
                            count += 1;
                        }
                    }
                    let cur_x = cur.0;
                    let cur_y = cur.1;
                    if count > 0 {
                        //if current's neighbour is Mine, update current to count of Mine
                        board[cur_x][cur_y] = ('0' as u8 + count) as char;
                    } else {
                        board[cur_x][cur_y] = 'B';
                        for d in &dir {
                            let x = cur.0 as i32 + d[0];
                            let y = cur.1 as i32 + d[1];
                            if x < 0 || y < 0 || x >= m || y >= n {
                                //check border case
                                continue;
                            }
                            if board[x as usize][y as usize] == 'E' {
                                board[x as usize][y as usize] = 'B';
                                //put new position to queue for next level
                                queue.push_back((x as usize, y as usize));
                            }
                        }
                    }
                }
            }
            board
        }
    }
  • 相关阅读:
    hmset
    java 调用mongo 判断大于等于 并且小约等于<=
    Maven项目,别人的没问题,自己机器一直有问题
    linux 时间datetimectl 问题
    真正手把手教 git
    0324-SQLMAP使用参数备注
    安全推荐网址:
    JavaScript Base64 作为文件上传的实例代码解析
    学习笔记|变量的解构赋值
    学习笔记|let 和 const 命令
  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/15773483.html
Copyright © 2011-2022 走看看