【力扣279. 完全平方数】完全背包+数学法-四平方和定理+BFS（python3）

题目描述

https://leetcode-cn.com/problems/perfect-squares/

思路题解

完全背包

一开始的思路：n=6665的时候，时间超限

class Solution:
    def numSquares(self, n: int) -> int:
        if n<4:return n
        dp=[10001]*(n+1)
        dp[1],dp[2],dp[3],dp[4]=1,2,3,1
        for i in range(5,n+1):
            if math.sqrt(i)%1!=0:
                for j in range(1,i//2+1):
                    dp[i]=min(dp[i-j]+dp[i-(i-j)],dp[i])
            else:dp[i]=1
        return dp[n]

看了下答案， 进行了改进

class Solution:
    def numSquares(self, n: int) -> int:
        if n<4:return n
        dp=[i for i in range(n+1)]
        for i in range(2,n+1):
            j=1
            while j*j<=i:
                dp[i]=min(dp[i-j*j],dp[i])
                j+=1
            dp[i]+=1
        return dp[n]

数学法-四平方和定理

https://leetcode-cn.com/problems/perfect-squares/solution/wan-quan-ping-fang-shu-by-leetcode-solut-t99c/

BFS：层序遍历树

https://leetcode-cn.com/problems/perfect-squares/solution/shu-ju-jie-gou-he-suan-fa-bfsdong-tai-gu-jl6u/

标签：perfect,力扣,cn,int,BFS,https,279,squares,leetcode
来源： https://blog.csdn.net/anan15151529/article/details/118308465