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Update 0062.不同路径.md
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```
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### Python
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递归
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```python
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class Solution: # 动态规划
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class Solution:
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def uniquePaths(self, m: int, n: int) -> int:
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dp = [[1 for i in range(n)] for j in range(m)]
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if m == 1 or n == 1:
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return 1
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return self.uniquePaths(m - 1, n) + self.uniquePaths(m, n - 1)
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```
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动态规划(版本一)
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```python
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class Solution:
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def uniquePaths(self, m: int, n: int) -> int:
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# 创建一个二维列表用于存储唯一路径数
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dp = [[0] * n for _ in range(m)]
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# 设置第一行和第一列的基本情况
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for i in range(m):
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dp[i][0] = 1
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for j in range(n):
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dp[0][j] = 1
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# 计算每个单元格的唯一路径数
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for i in range(1, m):
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for j in range(1, n):
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dp[i][j] = dp[i][j - 1] + dp[i - 1][j]
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return dp[m - 1][n - 1]
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```
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dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
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# 返回右下角单元格的唯一路径数
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return dp[m - 1][n - 1]
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```
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动态规划(版本二)
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```python
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class Solution:
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def uniquePaths(self, m: int, n: int) -> int:
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# 创建一个一维列表用于存储每列的唯一路径数
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dp = [1] * n
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# 计算每个单元格的唯一路径数
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for j in range(1, m):
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for i in range(1, n):
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dp[i] += dp[i - 1]
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# 返回右下角单元格的唯一路径数
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return dp[n - 1]
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```
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数论
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```python
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class Solution:
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def uniquePaths(self, m: int, n: int) -> int:
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numerator = 1 # 分子
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denominator = m - 1 # 分母
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count = m - 1 # 计数器,表示剩余需要计算的乘积项个数
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t = m + n - 2 # 初始乘积项
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while count > 0:
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numerator *= t # 计算乘积项的分子部分
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t -= 1 # 递减乘积项
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while denominator != 0 and numerator % denominator == 0:
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numerator //= denominator # 约简分子
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denominator -= 1 # 递减分母
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count -= 1 # 计数器减1,继续下一项的计算
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return numerator # 返回最终的唯一路径数
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```
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### Go
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```Go
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