## Description

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? Above is a 7 x 3 grid. How many possible unique paths are there?

Example 1:

Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right


Example 2:

Input: m = 7, n = 3
Output: 28


Constraints:

• 1 <= m, n <= 100
• It’s guaranteed that the answer will be less than or equal to 2 * 10 ^ 9.

## Solutions

寻找从开始到结束位置，有多少可能的唯一路径，每一步只能向右或者向下走。

### 1. DP

这一题还是很好求解的，只要列出状态转移方程就好了，用二维的 DP：

class Solution(object):
def uniquePaths(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
dp = [[0 for _ in range(n)] for _ in range(m)]
# dp = 1
for i in range(m):
for j in range(n):
if i == 0 and j != 0:
dp[i][j] = dp[i][j - 1]
elif i != 0 and j == 0:
dp[i][j] = dp[i - 1][j]
elif i == 0 and j == 0:
dp[i][j] = 1
else:
dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
return dp[m-1][n-1]

# Runtime: 20 ms, faster than 49.22% of Python online submissions for Unique Paths.
# Memory Usage: 11.9 MB, less than 6.90% of Python online submissions for Unique Paths.


然而这里有很多 if 的判断不是很清爽，可以在数组的每个维度上多加 1，然后改写一下就清爽很多了：

class Solution(object):
def uniquePaths(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
dp = 1
for i in range(1, m + 1):
for j in range(1, n + 1):
if i == 1 and j == 1:
continue
dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
return dp[m][n]

# Runtime: 20 ms, faster than 49.22% of Python online submissions for Unique Paths.
# Memory Usage: 11.8 MB, less than 31.03% of Python online submissions for Unique Paths.