Question

1730. 获取食物的最短路径

English Version

题目描述

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 '*' 。

解法

方法一：BFS

根据题意，我们需要从 * 出发，找到最近的 #，返回最短路径长度。

首先，我们遍历整个二维数组，找到 * 的位置，将其作为 BFS 的起点，放入队列中。

然后，我们开始 BFS，遍历队列中的元素，每次遍历到一个元素，我们将其上下左右四个方向的元素加入队列中，直到遇到 #，返回当前层数。

时间复杂度 $O(m \times n)$，空间复杂度 $O(1)$。其中 $m$ 和 $n$ 分别为二维数组的行数和列数。

class Solution:
  def getFood(self, grid: List[List[str]]) -> int:
    m, n = len(grid), len(grid[0])
    i, j = next((i, j) for i in range(m) for j in range(n) if grid[i][j] == '*')
    q = deque([(i, j)])
    dirs = (-1, 0, 1, 0, -1)
    ans = 0
    while q:
      ans += 1
      for _ in range(len(q)):
        i, j = q.popleft()
        for a, b in pairwise(dirs):
          x, y = i + a, j + b
          if 0 <= x < m and 0 <= y < n:
            if grid[x][y] == '#':
              return ans
            if grid[x][y] == 'O':
              grid[x][y] = 'X'
              q.append((x, y))
    return -1

class Solution {
  private int[] dirs = {-1, 0, 1, 0, -1};

  public int getFood(char[][] grid) {
    int m = grid.length, n = grid[0].length;
    Deque<int[]> q = new ArrayDeque<>();
    for (int i = 0, x = 1; i < m && x == 1; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] == '*') {
          q.offer(new int[] {i, j});
          x = 0;
          break;
        }
      }
    }
    int ans = 0;
    while (!q.isEmpty()) {
      ++ans;
      for (int t = q.size(); t > 0; --t) {
        var p = q.poll();
        for (int k = 0; k < 4; ++k) {
          int x = p[0] + dirs[k];
          int y = p[1] + dirs[k + 1];
          if (x >= 0 && x < m && y >= 0 && y < n) {
            if (grid[x][y] == '#') {
              return ans;
            }
            if (grid[x][y] == 'O') {
              grid[x][y] = 'X';
              q.offer(new int[] {x, y});
            }
          }
        }
      }
    }
    return -1;
  }
}

class Solution {
public:
  const static inline vector<int> dirs = {-1, 0, 1, 0, -1};

  int getFood(vector<vector<char>>& grid) {
    int m = grid.size(), n = grid[0].size();
    queue<pair<int, int>> q;
    for (int i = 0, x = 1; i < m && x == 1; ++i) {
      for (int j = 0; j < n; ++j) {
        if (grid[i][j] == '*') {
          q.emplace(i, j);
          x = 0;
          break;
        }
      }
    }
    int ans = 0;
    while (!q.empty()) {
      ++ans;
      for (int t = q.size(); t; --t) {
        auto [i, j] = q.front();
        q.pop();
        for (int k = 0; k < 4; ++k) {
          int x = i + dirs[k], y = j + dirs[k + 1];
          if (x >= 0 && x < m && y >= 0 && y < n) {
            if (grid[x][y] == '#') return ans;
            if (grid[x][y] == 'O') {
              grid[x][y] = 'X';
              q.emplace(x, y);
            }
          }
        }
      }
    }
    return -1;
  }
};

func getFood(grid [][]byte) (ans int) {
  m, n := len(grid), len(grid[0])
  dirs := []int{-1, 0, 1, 0, -1}
  type pair struct{ i, j int }
  q := []pair{}
  for i, x := 0, 1; i < m && x == 1; i++ {
    for j := 0; j < n; j++ {
      if grid[i][j] == '*' {
        q = append(q, pair{i, j})
        x = 0
        break
      }
    }
  }
  for len(q) > 0 {
    ans++
    for t := len(q); t > 0; t-- {
      p := q[0]
      q = q[1:]
      for k := 0; k < 4; k++ {
        x, y := p.i+dirs[k], p.j+dirs[k+1]
        if x >= 0 && x < m && y >= 0 && y < n {
          if grid[x][y] == '#' {
            return ans
          }
          if grid[x][y] == 'O' {
            grid[x][y] = 'X'
            q = append(q, pair{x, y})
          }
        }
      }
    }
  }
  return -1
}

/**
 * @param {character[][]} grid
 * @return {number}
 */
var getFood = function (grid) {
  const m = grid.length;
  const n = grid[0].length;
  const dirs = [-1, 0, 1, 0, -1];
  const q = [];
  for (let i = 0, x = 1; i < m && x == 1; ++i) {
    for (let j = 0; j < n; ++j) {
      if (grid[i][j] == '*') {
        q.push([i, j]);
        x = 0;
        break;
      }
    }
  }
  let ans = 0;
  while (q.length) {
    ++ans;
    for (let t = q.length; t > 0; --t) {
      const [i, j] = q.shift();
      for (let k = 0; k < 4; ++k) {
        const x = i + dirs[k];
        const y = j + dirs[k + 1];
        if (x >= 0 && x < m && y >= 0 && y < n) {
          if (grid[x][y] == '#') {
            return ans;
          }
          if (grid[x][y] == 'O') {
            grid[x][y] = 'X';
            q.push([x, y]);
          }
        }
      }
    }
  }
  return -1;
};