1730. 获取食物的最短路径（BFS）

1730. 获取食物的最短路径

你现在很饿，想要尽快找东西吃。你需要找到最短的路径到达一个食物所在的格子。

给定一个 m x n 的字符矩阵 grid ，包含下列不同类型的格子：

• '*' 是你的位置。矩阵中有且只有一个 '*' 格子。
• '#' 是食物。矩阵中可能存在多个食物。
• 'O' 是空地，你可以穿过这些格子。
• 'X' 是障碍，你不可以穿过这些格子。

返回你到任意食物的最短路径的长度。如果不存在你到任意食物的路径，返回 -1。

 

示例 1:

输入： grid = [["X","X","X","X","X","X"],["X","*","O","O","O","X"],["X","O","O","#","O","X"],["X","X","X","X","X","X"]]
输出： 3
解释： 要拿到食物，你需要走 3 步。

Example 2:

输入： grid = [["X","X","X","X","X"],["X","*","X","O","X"],["X","O","X","#","X"],["X","X","X","X","X"]]
输出： -1
解释： 你不可能拿到食物。

示例 3:

输入: grid = [["X","X","X","X","X","X","X","X"],["X","*","O","X","O","#","O","X"],["X","O","O","X","O","O","X","X"],["X","O","O","O","O","#","O","X"],["X","X","X","X","X","X","X","X"]]
输出: 6
解释: 这里有多个食物。拿到下边的食物仅需走 6 步。

 

提示：

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 200
• grid[row][col] 是 '*'、 'X'、 'O' 或 '#' 。
• grid 中有且只有一个 '*' 。

解题思路：

看到这个题目，很明显是一个BFS求最少步数问题

class Solution {
public:
    static constexpr int g_direction[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
    // 找到起始位置
    bool getStartPosition(const vector<vector<char>> &grid, std::pair<int, int> &start) {
        if (grid.size() <= 0 || grid[0].size() <= 0) {
            return false;
        }
        int row = grid.size();
        int cow = grid[0].size();
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < cow; j++) {
                if (grid[i][j] == '*') {
                    start = make_pair(i, j);
                    return true;
                }
            }
        }
        return false;
    }
    int getFood(vector<vector<char>>& grid) {
        if (grid.size() <= 0 || grid[0].size() <= 0) {
            return -1;
        }
        int row = grid.size();
        int cow = grid[0].size();
        // 找到起始位置('*'所在位置坐标)
        std::pair<int, int> start;
        if (getStartPosition(grid, start) != true) {
            return -1;
        }
        // 将起始位置入队列进行BFS,找到第一次到达食物位置的最短步数
        vector<vector<bool>> visited(row, vector<bool>(cow, false));
        queue<std::pair<int, int>> q;
        q.push(start);
        visited[start.first][start.second] = true;
        int step = 0;
        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                int x = q.front().first;
                int y = q.front().second;
                q.pop();
                // 如果找到食物,则返回走过的步数
                if (grid[x][y] == '#') {
                    return step;
                }
                // 如果没找到,则向四个方向继续找
                for (int j = 0; j < 4; j++) {
                    int xNext = x + g_direction[j][0];
                    int yNext = y + g_direction[j][1];
                    // 如果超过了边界或者遇到障碍物,则该格子不能走
                    if ((xNext >= 0 && xNext < row) && (yNext >= 0 && yNext < cow) && 
                        !visited[xNext][yNext] && grid[xNext][yNext] != 'X') {
                        q.push(make_pair(xNext, yNext));
                        visited[xNext][yNext] = true;
                    }
                }
            }
            step++;
        }
        return -1;
    }
};