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1091. 二进制矩阵中的最短路径

English Version

题目描述

给你一个 n x n 的二进制矩阵 grid 中，返回矩阵中最短 畅通路径 的长度。如果不存在这样的路径，返回 -1 。

二进制矩阵中的 畅通路径 是一条从 左上角 单元格（即，(0, 0)）到 右下角 单元格（即，(n - 1, n - 1)）的路径，该路径同时满足下述要求：

• 路径途经的所有单元格都的值都是 0 。
• 路径中所有相邻的单元格应当在 8 个方向之一 上连通（即，相邻两单元之间彼此不同且共享一条边或者一个角）。

畅通路径的长度 是该路径途经的单元格总数。

 

示例 1：

输入：grid = [[0,1],[1,0]]
输出：2

示例 2：

输入：grid = [[0,0,0],[1,1,0],[1,1,0]]
输出：4

示例 3：

输入：grid = [[1,0,0],[1,1,0],[1,1,0]]
输出：-1

 

提示：

• n == grid.length
• n == grid[i].length
• 1 <= n <= 100
• grid[i][j] 为 0 或 1

解法

BFS 最短路模型。

Python3

class Solution:
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
        if grid[0][0]:
            return -1
        n = len(grid)
        q = deque([(0, 0)])
        grid[0][0] = 1
        ans = 0
        while q:
            ans += 1
            for _ in range(len(q)):
                i, j = q.popleft()
                if (i, j) == (n - 1, n - 1):
                    return ans
                for x in range(i - 1, i + 2):
                    for y in range(j - 1, j + 2):
                        if 0 <= x < n and 0 <= y < n and grid[x][y] == 0:
                            q.append((x, y))
                            grid[x][y] = 1
        return -1

Java

class Solution {
    public int shortestPathBinaryMatrix(int[][] grid) {
        if (grid[0][0] == 1) {
            return -1;
        }
        int n = grid.length;
        Deque<int[]> q = new ArrayDeque<>();
        q.offer(new int[] {0, 0});
        grid[0][0] = 1;
        int ans = 0;
        while (!q.isEmpty()) {
            ++ans;
            for (int m = q.size(); m > 0; --m) {
                int[] p = q.poll();
                int i = p[0], j = p[1];
                if (i == n - 1 && j == n - 1) {
                    return ans;
                }
                for (int x = i - 1; x <= i + 1; ++x) {
                    for (int y = j - 1; y <= j + 1; ++y) {
                        if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 0) {
                            q.offer(new int[] {x, y});
                            grid[x][y] = 1;
                        }
                    }
                }
            }
        }
        return -1;
    }
}

C++

class Solution {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        if (grid[0][0]) return -1;
        int n = grid.size();
        queue<pair<int, int>> q;
        q.push({0, 0});
        grid[0][0] = 1;
        int ans = 0;
        while (!q.empty()) {
            ++ans;
            for (int m = q.size(); m > 0; --m) {
                auto p = q.front();
                q.pop();
                int i = p.first, j = p.second;
                if (i == n - 1 && j == n - 1) return ans;
                for (int x = i - 1; x <= i + 1; ++x) {
                    for (int y = j - 1; y <= j + 1; ++y) {
                        if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 0) {
                            q.push({x, y});
                            grid[x][y] = 1;
                        }
                    }
                }
            }
        }
        return -1;
    }
};

Go

func shortestPathBinaryMatrix(grid [][]int) int {
	if grid[0][0] == 1 {
		return -1
	}
	n := len(grid)
	q := [][]int{[]int{0, 0}}
	grid[0][0] = 1
	ans := 0
	for len(q) > 0 {
		ans++
		for m := len(q); m > 0; m-- {
			p := q[0]
			q = q[1:]
			i, j := p[0], p[1]
			if i == n-1 && j == n-1 {
				return ans
			}
			for x := i - 1; x <= i+1; x++ {
				for y := j - 1; y <= j+1; y++ {
					if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 0 {
						q = append(q, []int{x, y})
						grid[x][y] = 1
					}
				}
			}
		}
	}
	return -1
}

Rust

use std::collections::VecDeque;
impl Solution {
    pub fn shortest_path_binary_matrix(mut grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let mut queue = VecDeque::new();
        queue.push_back([0, 0]);
        let mut res = 0;
        while !queue.is_empty() {
            res += 1;
            for _ in 0..queue.len() {
                let [i, j] = queue.pop_front().unwrap();
                if grid[i][j] == 1 {
                    continue;
                }
                if i == n - 1 && j == n - 1 {
                    return res;
                }
                grid[i][j] = 1;
                for x in -1..=1 {
                    for y in -1..=1 {
                        let x = x + i as i32;
                        let y = y + j as i32;
                        if x < 0 || x == n as i32 || y < 0 || y == n as i32 {
                            continue;
                        }
                        queue.push_back([x as usize, y as usize]);
                    }
                }
            }
        }
        -1
    }
}

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