[Swift]LeetCode1091. 二进制矩阵中的最短路径 | Shortest Path in Binary Matrix

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/11014400.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

In an N by N square grid, each cell is either empty (0) or blocked (1).

A clear path from top-left to bottom-right has length k if and only if it is composed of cells C_1, C_2, ..., C_k such that:

• Adjacent cells C_i and C_{i+1} are connected 8-directionally (ie., they are different and share an edge or corner)
• C_1 is at location (0, 0) (ie. has value grid[0][0])
• C_k is at location (N-1, N-1) (ie. has value grid[N-1][N-1])
• If C_i is located at (r, c), then grid[r][c] is empty (ie. grid[r][c] == 0).

Return the length of the shortest such clear path from top-left to bottom-right.  If such a path does not exist, return -1. 

Example 1:

Input: [[0,1],[1,0]]
Output: 2

Example 2:

Input: [[0,0,0],[1,1,0],[1,1,0]]
Output: 4 

Note:

1. 1 <= grid.length == grid[0].length <= 100
2. grid[i][j] is 0 or 1

---

在一个 N × N 的方形网格中，每个单元格有两种状态：空（0）或者阻塞（1）。

一条从左上角到右下角、长度为 k 的畅通路径，由满足下述条件的单元格 C_1, C_2, ..., C_k 组成：

• 相邻单元格 C_i 和 C_{i+1} 在八个方向之一上连通（此时，C_i 和 C_{i+1} 不同且共享边或角）
• C_1 位于 (0, 0)（即，值为 grid[0][0]）
• C_k 位于 (N-1, N-1)（即，值为 grid[N-1][N-1]）
• 如果 C_i 位于 (r, c)，则 grid[r][c] 为空（即，grid[r][c] == 0）

返回这条从左上角到右下角的最短畅通路径的长度。如果不存在这样的路径，返回 -1 。

示例 1：

输入：[[0,1],[1,0]]

输出：2

示例 2：

输入：[[0,0,0],[1,1,0],[1,1,0]]

输出：4

提示：

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

---

Runtime: 516 ms

Memory Usage: 21 MB

class Solution {
    let dir:[[Int]] = [[0,1],[0,-1],[1,0],[-1,0],[1,-1],[-1,1],[-1,-1],[1,1]]
    func shortestPathBinaryMatrix(_ grid: [[Int]]) -> Int {
        let m:Int = grid.count
        let n:Int = grid[0].count
        
        if grid[0][0]==1 || grid[m-1][n-1]==1
        {
            return -1
        }
        
        var visited:[[Bool]] = [[Bool]](repeating:[Bool](repeating:false,count:n),count:m)
        visited[0][0] = true
        var queue:[[Int]] = [[Int]]()
        queue.append([0,0])
        var ans:Int = 0
        while(!queue.isEmpty)
        {
            let size:Int = queue.count
            for _ in 0..<size
            {
                var pop:[Int] = queue.removeFirst()
                if pop[0] == m-1 && pop[1] == n-1
                {
                    return ans + 1
                }
                for k in 0..<8
                {
                    let nextX:Int = dir[k][0]+pop[0]
                    let nextY:Int = dir[k][1]+pop[1]
                    if nextX>=0 && nextX<m && nextY>=0 && nextY<n && !visited[nextX][nextY] && grid[nextX][nextY]==0
                    {
                        queue.append([nextX,nextY])
                        visited[nextX][nextY] = true
                    }
                }
            }
            ans += 1
        }
        return -1
    }
}

---

612ms

class Solution {
    func shortestPathBinaryMatrix(_ grid: [[Int]]) -> Int {
        guard grid.count > 0 && grid[0].count > 0 else { return -1 }
        guard grid[0][0] == 0 else { return -1 }
        let R = grid.count
        let C = grid[0].count
        let directions = [(0, 1), (1, 0), (0, -1), (-1, 0), (-1, 1), (1, -1), (1, 1), (-1, -1)]
        var isVisited = [[Bool]](repeating: [Bool](repeating: false, count: grid[0].count), count: grid.count)
        var queue = [(0, 0)]
        isVisited[0][0] = true
        var steps = 1
        while queue.count > 0 {
            
            let size = queue.count
            steps += 1
            for i in 0..<size {
                let curr = queue.removeFirst()
                for direct in directions {
                    let nextR = curr.0 + direct.0
                    let nextC = curr.1 + direct.1
                    // print((nextR, nextC))
                    guard nextR < grid.count && nextR >= 0 && nextC < grid[0].count && nextC >= 0 && 
                         !isVisited[nextR][nextC] && grid[nextR][nextC] == 0 else {
                        continue
                    }
                    
                    // print((R, C))
                    if nextR == R-1 && nextC == C-1 { return steps }
                    queue.append((nextR, nextC))
                    isVisited[nextR][nextC] = true;
                }
            }
        }
        return -1
    }
}

---

868ms

class Solution {
    func shortestPathBinaryMatrix(_ grid: [[Int]]) -> Int {
        guard !grid.isEmpty, !grid[0].isEmpty else { return -1 }
        guard grid[0][0] == 0, grid[grid.count - 1][grid.count - 1] == 0 else { return -1 }
        var grid = grid
        
        let visiting = Set<[Int]>([[0, 0]])
        return bfs(&grid, visiting: visiting, level: 1)
    }
    
    private func bfs(_ grid: inout [[Int]], visiting: Set<[Int]>, level: Int) -> Int {
        guard grid[grid.count - 1][grid.count - 1] == 0 else { return level }
        
        let dirs: [[Int]] = [
            [-1, -1],
            [-1,  0],
            [-1,  1],
            [ 0,  1],
            [ 0, -1],
            [ 1, -1],
            [ 1,  0],
            [ 1,  1]
        ]
        var nextVisiting = Set<[Int]>()
        for p in visiting {
            dirs
            .map { [p[0] + $0[0], p[1] + $0[1]] }
            .forEach { np in
                if  np[0] >= 0, 
                    np[1] >= 0, 
                    np[0] < grid.count, 
                    np[1] < grid.count, 
                    grid[np[0]][np[1]] == 0 {
                    nextVisiting.insert(np)
                    grid[np[0]][np[1]] = 1
                }
            }
        }
        guard !nextVisiting.isEmpty else { return -1 }
        return bfs(&grid, visiting: nextVisiting, level: level + 1)
    }
}