From 36bdccd6878d0b54338741d699a4d7829318d1c1 Mon Sep 17 00:00:00 2001
From: Yuhao Ju <juguaguaji@gmail.com>
Date: Wed, 21 Dec 2022 17:17:59 +0800
Subject: [PATCH] =?UTF-8?q?update=200279.=E5=AE=8C=E5=85=A8=E5=B9=B3?=
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---
 problems/0279.完全平方数.md | 37 --------------------------------
 1 file changed, 37 deletions(-)

diff --git a/problems/0279.完全平方数.md b/problems/0279.完全平方数.md
index 50ddf5f9..b6a97be3 100644
--- a/problems/0279.完全平方数.md
+++ b/problems/0279.完全平方数.md
@@ -235,43 +235,6 @@ class Solution:
         return dp[n]
 ```
 
-Python3:
-```python
-class Solution:
-    '''版本一，先遍历背包, 再遍历物品'''
-    def numSquares(self, n: int) -> int:
-        dp = [n] * (n + 1)
-        dp[0] = 0
-        # 遍历背包
-        for j in range(1, n+1):
-            for i in range(1, n):
-                num = i ** 2
-                if num > j: break
-                # 遍历物品
-                if j - num >= 0:
-                    dp[j] = min(dp[j], dp[j - num] + 1)
-        return dp[n]
-
-class Solution:
-    '''版本二， 先遍历物品, 再遍历背包'''
-    def numSquares(self, n: int) -> int:
-        # 初始化
-        # 组成和的完全平方数的最多个数，就是只用1构成
-        # 因此，dp[i] = i
-        dp = [i for i in range(n + 1)]
-        # dp[0] = 0 无意义，只是为了方便记录特殊情况:
-        # n本身就是完全平方数，dp[n] = min(dp[n], dp[n - n] + 1) = 1
-
-        for i in range(1, n): # 遍历物品
-            if i * i > n:
-                break
-            num = i * i
-            for j in range(num, n + 1): # 遍历背包
-                dp[j] = min(dp[j], dp[j - num] + 1)
-
-        return dp[n]
-```
-
 Go：
 ```go
 // 版本一,先遍历物品, 再遍历背包