題目在此 1091. Shortest Path in Binary Matrix
給定一個地圖找出最短路徑
![image 25]()
解題思維
這題很直覺的可以用 Breadth-First Search 來解決問題
一摸到終點就結束了
程式碼
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| class Solution:
def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
def get_neighbors(y, x):
offset = [-1, 0, 1]
result = []
for offset_y in offset:
new_y = y + offset_y
if not 0 <= new_y < length:
continue
for offset_x in offset:
if offset_y == 0 and offset_x == 0:
continue
new_x = x + offset_x
if not (0 <= new_x < length):
continue
if grid[new_y][new_x] == 1:
continue
if path_map[new_y][new_x] != -1:
continue
result.append((new_y, new_x))
return result
length = len(grid[0])
if length <= 1:
return length
if grid[0][0] == 1 or grid[length - 1][length - 1] == 1:
return -1
path_map = [[-1 for _ in range(length)] for _ in range(length)]
path_map[0][0] = 1
changed_0 = [(0, 0)]
changed_1 = []
while changed_0:
for y, x in changed_0:
neighbors = get_neighbors(y, x)
new_result = path_map[y][x] + 1
if (length - 1, length - 1) in neighbors:
return new_result
for n_y, n_x in neighbors:
path_map[n_y][n_x] = new_result
changed_1.extend(neighbors)
changed_0 = changed_1
changed_1 = []
return -1
|
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