LeetCode //C - 1926. Nearest Exit from Entrance in Maze

1926. Nearest Exit from Entrance in Maze

You are given an m x n matrix maze (0-indexed) with empty cells (represented as '.') and walls (represented as '+'). You are also given the entrance of the maze, where entrance = [entrancerow, entrancecol] denotes the row and column of the cell you are initially standing at.

In one step, you can move one cell up, down, left, or right. You cannot step into a cell with a wall, and you cannot step outside the maze. Your goal is to find the nearest exit from the entrance. An exit is defined as an empty cell that is at the border of the maze. The entrance does not count as an exit.

Return the number of steps in the shortest path from the entrance to the nearest exit, or -1 if no such path exists.

Example 1:

Input: maze = [["+","+",".","+"],[".",".",".","+"],["+","+","+","."]], entrance = [1,2]
Output: 1
Explanation: There are 3 exits in this maze at [1,0], [0,2], and [2,3].

Initially, you are at the entrance cell [1,2].

  • You can reach [1,0] by moving 2 steps left.

  • You can reach [0,2] by moving 1 step up.

It is impossible to reach [2,3] from the entrance.

Thus, the nearest exit is [0,2], which is 1 step away.

Example 2:

Input: maze = [["+","+","+"],[".",".","."],["+","+","+"]], entrance = [1,0]
Output: 2
Explanation: There is 1 exit in this maze at [1,2].

1,0\] does not count as an exit since it is the entrance cell. Initially, you are at the entrance cell \[1,0\]. - You can reach \[1,2\] by moving 2 steps right. Thus, the nearest exit is \[1,2\], which is 2 steps away.

Example 3:

Input: maze = [[".","+"]], entrance = [0,0]
Output: -1
Explanation: There are no exits in this maze.

Constraints:
  • maze.length == m
  • maze[i].length == n
  • 1 <= m, n <= 100
  • maze[i][j] is either '.' or '+'.
  • entrance.length == 2
  • 0 <= entrancerow < m
  • 0 <= entrancecol < n
  • entrance will always be an empty cell.

From: LeetCode

Link: 1926. Nearest Exit from Entrance in Maze


Solution:

Ideas:

This function uses Breadth-First Search (BFS) to find the nearest exit. It checks all possible paths from the entrance and returns the number of steps to the nearest exit. If no exit is reachable, it returns -1. Please note that this code should be part of a complete C program and compiled with a C compiler to run. It assumes that the maze is properly allocated and passed to the function along with the size parameters and entrance coordinates.

Code:
c 复制代码
// Helper function to check if the current position is valid and not a wall.
int isValid(char **maze, int x, int y, int m, int n) {
    if (x >= 0 && x < m && y >= 0 && y < n && maze[x][y] == '.')
        return 1;
    return 0;
}

// Helper function to check if the current position is at the border and not the entrance.
int isExit(char **maze, int x, int y, int m, int n, int* entrance) {
    if ((x == 0 || x == m - 1 || y == 0 || y == n - 1) && (x != entrance[0] || y != entrance[1]))
        return 1;
    return 0;
}

int nearestExit(char** maze, int mazeSize, int* mazeColSize, int* entrance, int entranceSize) {
    int directions[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; // Up, Down, Left, Right
    int m = mazeSize, n = mazeColSize[0];
    int queueSize = m * n;
    int queue[queueSize][3]; // Queue to store the x, y, and distance
    int front = 0, rear = 0;

    // Initialize the queue with the entrance position
    queue[rear][0] = entrance[0];
    queue[rear][1] = entrance[1];
    queue[rear][2] = 0;
    rear++;

    maze[entrance[0]][entrance[1]] = '+'; // Mark the entrance as visited

    while (front < rear) {
        // Dequeue the front element
        int x = queue[front][0];
        int y = queue[front][1];
        int dist = queue[front][2];
        front++;

        // Check if we've reached an exit
        if (isExit(maze, x, y, m, n, entrance)) {
            return dist;
        }

        // Check all four possible directions
        for (int i = 0; i < 4; i++) {
            int newX = x + directions[i][0];
            int newY = y + directions[i][1];

            // If valid, not a wall, and not visited, add to queue
            if (isValid(maze, newX, newY, m, n)) {
                maze[newX][newY] = '+'; // Mark as visited
                queue[rear][0] = newX;
                queue[rear][1] = newY;
                queue[rear][2] = dist + 1;
                rear++;
            }
        }
    }

    // If no exit is reached, return -1
    return -1;
}
相关推荐
AAA.建材批发刘哥6 分钟前
03--C++ 类和对象中篇
linux·c语言·开发语言·c++·经验分享
小O的算法实验室26 分钟前
2024年IEEE TMC SCI1区TOP,面向无人机辅助 MEC 系统的轨迹规划与任务卸载的双蚁群算法,深度解析+性能实测
算法·无人机·论文复现·智能算法·智能算法改进
无才顽石1 小时前
什么是数学
算法·数理象
AlexMercer10121 小时前
【操作系统】操作系统期末考试 简答题 焚决
c语言·经验分享·笔记·操作系统
CoderCodingNo1 小时前
【GESP】C++五级真题(数论, 贪心思想考点) luogu-B4070 [GESP202412 五级] 奇妙数字
开发语言·c++·算法
百***58841 小时前
MATLAB高效算法实战技术文章大纲1
人工智能·算法·matlab
学嵌入式的六子1 小时前
如何使用VScode开发STM32【喂饭级教程】-全过程讲解
c语言·ide·vscode·stm32·单片机·嵌入式硬件
墨辰JC2 小时前
C语言可变参数讲解:stdarg.h应用
c语言·开发语言·蓝桥杯·内存·蓝桥杯嵌入式
C语言小火车2 小时前
C++右值引用与转移语义详解
c语言·开发语言
hans汉斯2 小时前
【人工智能与机器人研究】自动移液设备多轴运动控制系统设计
算法·机器学习·3d·自然语言处理·机器人·硬件架构·汉斯出版社