【leetcode】1091. Shortest Path in Binary Matrix
题目如下:
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 cellsC_1, C_2, ..., C_k
such that:
- Adjacent cells
C_i
andC_{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 valuegrid[0][0]
)C_k
is at location(N-1, N-1)
(ie. has valuegrid[N-1][N-1]
)- If
C_i
is located at(r, c)
, thengrid[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: 2Example 2:
Input: [[0,0,0],[1,1,0],[1,1,0]]
Output: 4Note:
1 <= grid.length == grid[0].length <= 100
grid[r][c]
is0
or1
解题思路:典型的BFS问题。从起点开始依次计算每个方格到达的最小值即可。这里用visit[i][j]记录从起点开始到(i,j)的最短路径,在BFS的过程中,可能有多条路径都能到达(i,j),因此只要有更短的走法能到达(i,j),就需要把(i,j)重新加入到队列中。
代码如下:
class Solution(object):
def shortestPathBinaryMatrix(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
if grid[0][0] == 1:
return -1
visit = []
val = []
for i in grid:
visit.append([0] * len(i))
val.append([0] * len(i))
visit[0][0] = 1
queue = [(0,0)]
direction = [(0,1),(0,-1),(1,0),(1,-1),(-1,1),(-1,-1),(1,1),(1,-1)]
while len(queue) > 0:
x,y = queue.pop(0)
for (i,j) in direction:
if x + i >= 0 and x+i < len(grid) and y + j >= 0 and y + j < len(grid[0]) \
and grid[x + i][y+j] == 0 and (visit[x+i][y+j] == 0 or visit[x+i][y+j] - 1 > visit[x][y]):
queue.append((x+i,y+j))
visit[x+i][y+j] = visit[x][y] + 1
return visit[-1][-1] if visit[-1][-1] > 0 else -1
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