124.Perfect Squares

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

.

Credits:

Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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给定一个正整数，求该正整数可以被被分解成完全平方数的个数。

For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.

采用动态规划的思想。

如果n为完全平方数，则f
=1;

否则f(n) = min{f(k)+f(n-k)},1<=k<=n/2

/**
* 给定一个正整数，求该正整数可以被被分解成完全平方数的个数。
* For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.
* 采用动态规划的思想。
* 如果n为完全平方数，则f
=1;
* 否则f(n) = min{f(k)+f(n-k)},1<=k<=n/2
* @date 20160510
* @param n
* @return
*/
public int numSquares(int n) {

if(n == 1){
return 1;
}
int f[] = new int[n+1];//f[i]表示给定i时其对应的符合条件的完全平方数个数。
f[1]=1;

for(int i=2;i<=n;i++){
/*首先判断是否为一个完全平方数*/
double y = Math.sqrt(i);
if( y - (int)y == 0){
f[i] = 1;
continue;
}
/*不为完全平方数时*/
f[i] = f[1]+f[i-1];
for(int k=2;k<=i/2;k++){
if(f[k]+f[i-k]<f[i]){
f[i] = f[k]+f[i-k];
}
}
}
return f
;
}