难度: Medium
原题连接
内容描述
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
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.
Note: m and n will be at most 100.
Example 1:
Input:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right
思路 1 - 时间复杂度: O(M * N)- 空间复杂度: O(1)******
递归+DFS
有个很恶心的case，就是obstacleGrid[0][0] == 1，这样怎么都走不到右下角
但是超时了
class Solution(object):
def uniquePathsWithObstacles(self, obstacleGrid):
"""
:type obstacleGrid: List[List[int]]
:rtype: int
"""
self.res = 0
def travel(x, y):
if x == len(obstacleGrid) - 1 and y == len(obstacleGrid[0]) - 1 and obstacleGrid[x][y] != 1:
self.res += 1
if x + 1 <= len(obstacleGrid) - 1 and obstacleGrid[x+1][y] != 1:
travel(x+1, y)
if y + 1 <= len(obstacleGrid[0]) - 1 and obstacleGrid[x][y+1] != 1:
travel(x, y+1)
if obstacleGrid[0][0] == 1:
return 0
travel(0, 0)
return self.res
思路 2 - 时间复杂度: O(M * N)- 空间复杂度: O(M * N)******
加一个cache
beats 66.49%
class Solution(object):
def uniquePathsWithObstacles(self, obstacleGrid):
"""
:type obstacleGrid: List[List[int]]
:rtype: int
"""
cache = {}
def travel(x, y):
if (x, y) in cache:
return cache[x, y]
res = 0
if obstacleGrid[x][y] == 0:
if x + 1 <= len(obstacleGrid) - 1:
res += travel(x+1, y)
if y + 1<= len(obstacleGrid[0]) - 1:
res += travel(x, y+1)
if x == len(obstacleGrid) - 1 and y == len(obstacleGrid[0]) - 1:
res += 1
cache[x, y] = res
return cache[x, y]
if obstacleGrid[0][0] == 1:
return 0
return travel(0, 0)思路 3 - 时间复杂度: O(M * N)- 空间复杂度: O(M * N)******
DP， beats 100%
class Solution(object):
def uniquePathsWithObstacles(self, obstacleGrid):
"""
:type obstacleGrid: List[List[int]]
:rtype: int
"""
m = len(obstacleGrid)
n = len(obstacleGrid[0])
# If the starting cell has an obstacle, then simply return as there would be
# no paths to the destination.
if obstacleGrid[0][0] == 1:
return 0
# Number of ways of reaching the starting cell = 1.
obstacleGrid[0][0] = 1
# Filling the values for the first column
for i in range(1, m):
obstacleGrid[i][0] = int(obstacleGrid[i][0] == 0 and obstacleGrid[i-1][0] == 1)
# Filling the values for the first row
for j in range(1, n):
obstacleGrid[0][j] = int(obstacleGrid[0][j] == 0 and obstacleGrid[0][j-1] == 1)
# Starting from cell(1,1) fill up the values
# No. of ways of reaching cell[i][j] = cell[i - 1][j] + cell[i][j - 1]
# i.e. From above and left.
for i in range(1, m):
for j in range(1, n):
if obstacleGrid[i][j] == 0:
obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1]
else:
obstacleGrid[i][j] = 0
# Return value stored in rightmost bottommost cell. That is the destination.
return obstacleGrid[m-1][n-1]