LeetCode //C - 130. Surrounded Regions

130. Surrounded Regions

Given an m x n matrix board containing 'X' and 'O' , capture all regions that are 4-directionally surrounded by 'X'.

A region is captured by flipping all 'O' s into 'X' s in that surrounded region.

Example 1:

Input: board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
Output: [["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
Explanation: Notice that an 'O' should not be flipped if:

  • It is on the border, or
  • It is adjacent to an 'O' that should not be flipped.

The bottom 'O' is on the border, so it is not flipped.

The other three 'O' form a surrounded region, so they are flipped.

Example 2:

Input: board = [["X"]]
Output: [["X"]]

Constraints:
  • m == board.length
  • n == board[i].length
  • 1 <= m, n <= 200
  • board[i][j] is 'X' or 'O'.

From: LeetCode

Link: 130. Surrounded Regions


Solution:

Ideas:
  1. Iterate over the boundary (four sides) of the board.
  2. For every 'O' on the boundary, perform a Depth First Search (DFS) to mark all 'O's connected to it with a temporary marker, such as 'B', to denote that these 'O's are on the boundary or connected to the boundary and should not be flipped.
  3. After marking all 'O's on the boundary and the ones connected to it, iterate over the entire board. Perform the following operations:
    • Change all 'B' to 'O' as these are not surrounded by 'X'.
    • Change all remaining 'O' to 'X' as these are surrounded by 'X'.
Code:
c 复制代码
void dfs(char** board, int i, int j, int m, int n) {
    if(i < 0 || i >= m || j < 0 || j >= n || board[i][j] != 'O') return;
    
    // mark the current cell as 'B'
    board[i][j] = 'B';
    
    // perform DFS in all four directions
    dfs(board, i+1, j, m, n);
    dfs(board, i-1, j, m, n);
    dfs(board, i, j+1, m, n);
    dfs(board, i, j-1, m, n);
}

void solve(char** board, int boardSize, int* boardColSize) {
    if(boardSize == 0 || boardColSize[0] == 0) return;
    
    int m = boardSize, n = boardColSize[0];
    
    // Step 1: mark the boundary 'O's and the ones connected to them with 'B'
    for(int i = 0; i < m; i++) {
        dfs(board, i, 0, m, n); // left boundary
        dfs(board, i, n-1, m, n); // right boundary
    }
    
    for(int j = 0; j < n; j++) {
        dfs(board, 0, j, m, n); // top boundary
        dfs(board, m-1, j, m, n); // bottom boundary
    }
    
    // Step 2: flip the remaining 'O's to 'X' and 'B's back to 'O'
    for(int i = 0; i < m; i++) {
        for(int j = 0; j < n; j++) {
            if(board[i][j] == 'O') board[i][j] = 'X';
            if(board[i][j] == 'B') board[i][j] = 'O';
        }
    }
}
相关推荐
violet-lz18 分钟前
数据结构四大简单排序算法详解:直接插入排序、选择排序、基数排序和冒泡排序
数据结构·算法·排序算法
·白小白22 分钟前
力扣(LeetCode) ——118.杨辉三角(C++)
c++·算法·leetcode
CoovallyAIHub1 小时前
超越“识别”:下一代机器视觉如何破解具身智能落地难题?
深度学习·算法·计算机视觉
仰泳的熊猫1 小时前
LeetCode:207. 课程表
数据结构·c++·算法·leetcode
liu****1 小时前
19.map和set的封装
开发语言·数据结构·c++·算法
水冗水孚1 小时前
双指针算法在实际开发中的具体应用之代码Review文章字符串的片段分割
算法·leetcode
DuHz1 小时前
用于汽车雷达应用的步进频率PMCW波形——论文阅读
论文阅读·算法·汽车·信息与通信·信号处理·毫米波雷达
润 下1 小时前
C语言——回调函数的典型示例(分析详解)
c语言·开发语言·人工智能·经验分享·笔记·程序人生
张晓~183399481211 小时前
碰一碰发抖音源码技术搭建部署方案
线性代数·算法·microsoft·矩阵·html5
weixin_448119941 小时前
Datawhale人工智能的数学基础 202510第3次作业
人工智能·算法