[279]Perfect Squares

【题目描述】

Given a positive integer n, find the least number of perfect square numbers (for example, 

1,
4, 9, 16, ...

) which sum to n.

For example, given n = 

12

, return 

3

 because 

12
= 4 + 4 + 4

; given n = 

13

, return 

2

 because 

13
= 4 + 9

.
【思路】
1.用动态规划。f[i]=min(f[i],f[i-j*j])

2.根据四平方和定理和https://leetcode.com/discuss/56982/o-sqrt-n-in-ruby-and-c，

... a square if and only if each prime factor occurs to an even power in the number's prime factorization.
... a sum of two squares if
and only if each prime factor that's 3 modulo 4 occurs to an even power in the number's prime factorization.
... a sum of three squares if
and only if it's not of the form 4a(8b+7) with integers a and b.
... a sum of four squares.
Period. No condition. You never need more than four。

【代码】

<span style="font-size:14px;">class Solution {
public:
int numSquares(int n) {
vector<int> f(n+1,INT_MAX);
for(int i=1;i*i<=n;i++){
f[i*i]=1;
}
for(int i=2;i<=n;i++){
if(f[i]==1) continue;
for(int j=1;j*j<i;j++){
f[i]=min(f[i],f[i-j*j]+1);
}
}
return f
;
}
};</span>

class Solution {
public:
int numSquares(int n) {
while(n%4==0) n/=4;
if(n%8==7) return 4;
for(int i=0;i*i<=n;i++){
int j=sqrt(n-i*i);
if(i*i+j*j==n){
return !!i+!!j;
}
}
return 3;
}
};