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.

class Solution {
    public int numSquares(int n) {
        int[] dp = new int[n+1];//數字i能分解的個數
        for(int i=1; i<=n; i++){
            int j=1;
            int min=Integer.MAX_VALUE;
            while(i-j*j>=0){
                min = Math.min(min, dp[i-j*j]+1);
                j++;
            }     
            dp[i] = min;
        }
        return dp[n];
    }  
}