From 6128c57e151b55d433ee74135838c303a9694908 Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Thu, 1 Jun 2023 17:13:23 -0500
Subject: [PATCH] =?UTF-8?q?Update=200062.=E4=B8=8D=E5=90=8C=E8=B7=AF?=
 =?UTF-8?q?=E5=BE=84.md?=
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---
 problems/0062.不同路径.md | 63 ++++++++++++++++++++++++++++++++---
 1 file changed, 58 insertions(+), 5 deletions(-)

diff --git a/problems/0062.不同路径.md b/problems/0062.不同路径.md
index bf436369..430dacdc 100644
--- a/problems/0062.不同路径.md
+++ b/problems/0062.不同路径.md
@@ -287,17 +287,70 @@ public:
 ```
 
 ### Python
-
+递归
 ```python
-class Solution: # 动态规划
+class Solution:
     def uniquePaths(self, m: int, n: int) -> int:
-        dp = [[1 for i in range(n)] for j in range(m)]
+        if m == 1 or n == 1:
+            return 1
+        return self.uniquePaths(m - 1, n) + self.uniquePaths(m, n - 1)
+
+```
+动态规划（版本一）
+```python
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        # 创建一个二维列表用于存储唯一路径数
+        dp = [[0] * n for _ in range(m)]
+        
+        # 设置第一行和第一列的基本情况
+        for i in range(m):
+            dp[i][0] = 1
+        for j in range(n):
+            dp[0][j] = 1
+        
+        # 计算每个单元格的唯一路径数
         for i in range(1, m):
             for j in range(1, n):
-                dp[i][j] = dp[i][j - 1] + dp[i - 1][j]
+                dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
+        
+        # 返回右下角单元格的唯一路径数
         return dp[m - 1][n - 1]
-```
 
+```
+动态规划（版本二）
+```python
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        # 创建一个一维列表用于存储每列的唯一路径数
+        dp = [1] * n
+        
+        # 计算每个单元格的唯一路径数
+        for j in range(1, m):
+            for i in range(1, n):
+                dp[i] += dp[i - 1]
+        
+        # 返回右下角单元格的唯一路径数
+        return dp[n - 1]
+```
+数论
+```python
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        numerator = 1  # 分子
+        denominator = m - 1  # 分母
+        count = m - 1  # 计数器，表示剩余需要计算的乘积项个数
+        t = m + n - 2  # 初始乘积项
+        while count > 0:
+            numerator *= t  # 计算乘积项的分子部分
+            t -= 1  # 递减乘积项
+            while denominator != 0 and numerator % denominator == 0:
+                numerator //= denominator  # 约简分子
+                denominator -= 1  # 递减分母
+            count -= 1  # 计数器减1，继续下一项的计算
+        return numerator  # 返回最终的唯一路径数
+
+```
 ### Go
 
 ```Go