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
沐怡旸19 小时前
【算法--链表】114.二叉树展开为链表--通俗讲解
算法·面试
一只懒洋洋19 小时前
K-meas 聚类、KNN算法、决策树、随机森林
算法·决策树·聚类
小莞尔19 小时前
【51单片机】【protues仿真】基于51单片机停车场的车位管理系统
c语言·开发语言·单片机·嵌入式硬件·51单片机
xianyinsuifeng20 小时前
Oracle 10g → Oracle 19c 升级后问题解决方案(Pro*C 项目)
c语言·数据库·oracle
方案开发PCBA抄板芯片解密20 小时前
什么是算法:高效解决问题的逻辑框架
算法
songx_9920 小时前
leetcode9(跳跃游戏)
数据结构·算法·游戏
学c语言的枫子20 小时前
数据结构——双向链表
c语言·数据结构·链表
小白狮ww21 小时前
RStudio 教程:以抑郁量表测评数据分析为例
人工智能·算法·机器学习
AAA修煤气灶刘哥21 小时前
接口又被冲崩了?Sentinel 这 4 种限流算法,帮你守住后端『流量安全阀』
后端·算法·spring cloud
kk”1 天前
C语言快速排序
数据结构·算法·排序算法