From 5e68f30728cbcdb670eda7f12d3c208796f5c731 Mon Sep 17 00:00:00 2001 From: Jijie LIU Date: Sun, 27 Jun 2021 17:40:11 +0200 Subject: [PATCH] =?UTF-8?q?update=20279.=E5=AE=8C=E5=85=A8=E5=B9=B3?= =?UTF-8?q?=E6=96=B9=E6=95=B0:=20=E6=8F=90=E4=BE=9B=E6=96=B0=E7=9A=84?= =?UTF-8?q?=E5=88=9D=E5=A7=8B=E5=8C=96=E6=96=B9=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0279.完全平方数.md | 18 ++++++++++++++++++ 1 file changed, 18 insertions(+) diff --git a/problems/0279.完全平方数.md b/problems/0279.完全平方数.md index 60d6d165..d0922de1 100644 --- a/problems/0279.完全平方数.md +++ b/problems/0279.完全平方数.md @@ -214,8 +214,26 @@ class Solution: return dp[n] ``` +Python3: +```python +class Solution: + 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