1091. Shortest Path in Binary Matrix

# 1091. Shortest Path in Binary Matrix#

## 题目 #

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

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[r][c]` is `0` or `1`

## 题目大意 #

• 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）

## 解题思路 #

• 这一题是简单的找最短路径。利用 BFS 从左上角逐步扩展到右下角，便可以很容易求解。注意每轮扩展需要考虑 8 个方向。

## 代码 #

``````var dir = [][]int{
{-1, -1},
{-1, 0},
{-1, 1},
{0, 1},
{0, -1},
{1, -1},
{1, 0},
{1, 1},
}

func shortestPathBinaryMatrix(grid [][]int) int {
visited := make([][]bool, 0)
for range make([]int, len(grid)) {
visited = append(visited, make([]bool, len(grid[0])))
}
dis := make([][]int, 0)
for range make([]int, len(grid)) {
dis = append(dis, make([]int, len(grid[0])))
}
if grid[0][0] == 1 {
return -1
}
if len(grid) == 1 && len(grid[0]) == 1 {
return 1
}

queue := []int{0}
visited[0][0], dis[0][0] = true, 1
for len(queue) > 0 {
cur := queue[0]
queue = queue[1:]
curx, cury := cur/len(grid[0]), cur%len(grid[0])
for d := 0; d < 8; d++ {
nextx := curx + dir[d][0]
nexty := cury + dir[d][1]
if isInBoard(grid, nextx, nexty) && !visited[nextx][nexty] && grid[nextx][nexty] == 0 {
queue = append(queue, nextx*len(grid[0])+nexty)
visited[nextx][nexty] = true
dis[nextx][nexty] = dis[curx][cury] + 1
if nextx == len(grid)-1 && nexty == len(grid[0])-1 {
return dis[nextx][nexty]
}
}
}
}
return -1
}

func isInBoard(board [][]int, x, y int) bool {
return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
}
``````

Apr 8, 2023