题目地址(130. 被围绕的区域)

Posted 2022-05-24 潜行前行

题目地址(130. 被围绕的区域)

https://leetcode.cn/problems/surrounded-regions/

题目描述

给你一个 m x n 的矩阵 board ，由若干字符 'X' 和 'O' ，找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。

 

示例 1：

输入：board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
输出：[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
解释：被围绕的区间不会存在于边界上，换句话说，任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上，或不与边界上的 'O' 相连的 'O' 最终都会被填充为 'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。


示例 2：

输入：board = [["X"]]
输出：[["X"]]


 

提示：

m == board.length
n == board[i].length
1 <= m, n <= 200
board[i][j] 为 'X' 或 'O'

关键点

• 记忆搜索

代码

• 语言支持：Java

Java Code:

class Solution 
    int[][] dir = new int[][]1,0,-1,0,0,1,0,-1;
    public void solve(char[][] board) 
        
        for(int i = 0; i < board.length;i++ )
            for(int j = 0; j < board[0].length;j++ )
                // 从边缘开始遍历
                if( i ==0 || j == 0 || i == (board.length-1) || j == (board[0].length-1) )
                    if(board[i][j] == 'O')
                        dfs(i,j,board);
                
            
        
        for(int i = 0; i < board.length;i++ )
            for(int j = 0; j < board[0].length;j++ )
                if(board[i][j] == 'O')
                    board[i][j] = 'X';
                else if(board[i][j] == '1')
                    board[i][j] = 'O';
                
            
        
    

    void dfs(int i,int j,char[][] board)
        if( i < 0 || j < 0 || i > (board.length-1) || j > (board[0].length-1) || board[i][j] == 'X' ||  board[i][j] == '1' )
            return;
        board[i][j] = '1';
        for(int[] epoch : dir)
            dfs(i+epoch[0],j+epoch[1],board);
        
    

复杂度分析

令 n 为数组长度。

• 时间复杂度：$O(n)$
• 空间复杂度：$O(n)$

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