-
Notifications
You must be signed in to change notification settings - Fork 20
/
UniquePathsII.py
executable file
·58 lines (51 loc) · 1.55 KB
/
UniquePathsII.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
# -*- coding: UTF-8 -*-
#
# Follow up for "Unique Paths":
# Now consider if some obstacles are added to the grids. How many unique paths would there be?
# An obstacle and empty space is marked as 1 and 0 respectively in the grid.
#
# For example,
# There is one obstacle in the middle of a 3x3 grid as illustrated below.
#
# [
# [0,0,0],
# [0,1,0],
# [0,0,0]
# ]
#
# The total number of unique paths is 2.
#
# Note: m and n will be at most 100.
#
# Python, Python 3 all accepted.
class UniquePathsII:
def uniquePathsWithObstacles(self, obstacleGrid):
"""
:type obstacleGrid: List[List[int]]
:rtype: int
"""
if obstacleGrid[0][0] == 1:
return 0
if len(obstacleGrid) == 1:
for i in obstacleGrid[0]:
if i == 1:
return 0
return 1
matrix = [[0 for x in range(len(obstacleGrid[0]))] for x in range(len(obstacleGrid))]
for i in range(len(obstacleGrid)):
if obstacleGrid[i][0] == 1:
break
else:
matrix[i][0] = 1
for i in range(len(obstacleGrid[0])):
if obstacleGrid[0][i] == 1:
break
else:
matrix[0][i] = 1
for i in range(1, len(matrix)):
for j in range(1, len(matrix[0])):
if obstacleGrid[i][j] == 1:
matrix[i][j] = 0
else:
matrix[i][j] = matrix[i - 1][j] + matrix[i][j - 1]
return matrix[len(matrix) - 1][len(matrix[0]) - 1]