LeetCode 62 & 63: Unique Paths
Unique Paths I
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).
How many possible unique paths are there?
https://leetcode.com/problems/unique-paths/description/
class Solution:
def uniquePaths(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
# 即求组合C(m+n-2, n-1)
p = m+n-2
q = n-1
U = 1
D = 1
for i in range(1, q+1):
U *= p
D *= i
p -= 1
return U//D
Unique Paths II
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
https://leetcode.com/problems/unique-paths-ii/description/
class Solution:
def uniquePathsWithObstacles(self, obstacleGrid):
"""
:type obstacleGrid: List[List[int]]
:rtype: int
"""
# DP
R = len(obstacleGrid)
C = len(obstacleGrid[0])
arr = [[0] * C for _ in range(R)]
if obstacleGrid[0][0] == 0:
arr[0][0] = 1
else:
arr[0][0] = 0
for i in range(1, R):
if obstacleGrid[i][0] == 0:
arr[i][0] = arr[i-1][0]
else:
arr[i][0] = 0
for j in range(1, C):
if obstacleGrid[0][j] == 0:
arr[0][j] = arr[0][j-1]
else:
arr[0][j] = 0
for i in range(1, R):
for j in range(1, C):
if obstacleGrid[i][j] == 0:
# 或从上面过来, 或从左面过来
arr[i][j] = arr[i-1][j] + arr[i][j-1]
else:
arr[i][j] = 0
return arr[-1][-1]
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