<p>Given an integer <code>n</code>, return <em>the least number of perfect square numbers that sum to</em> <code>n</code>.</p>

<p>A <strong>perfect square</strong> is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, <code>1</code>, <code>4</code>, <code>9</code>, and <code>16</code> are perfect squares while <code>3</code> and <code>11</code> are not.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 12
<strong>Output:</strong> 3
<strong>Explanation:</strong> 12 = 4 + 4 + 4.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 13
<strong>Output:</strong> 2
<strong>Explanation:</strong> 13 = 4 + 9.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int* table;
    int* ceilingSqr;
    int* squares;
    void recur(int n){
        table[n]=n-1;
        for(int i=ceilingSqr[n];i>-1;--i){
            table[n]=min(table[n], table[n-squares[i]]);
            if(table[n]==1) break;
        }
        table[n]++;
    }
    int numSquares(int n) {
        int sz=(n+1)<<2;
        int tableLength=n+1;
        int sqroot=int(sqrt(n));
        table=new int[tableLength];
        ceilingSqr=new int[tableLength];
        memset(table, 0, sz);
        squares=new int[sqroot];

        //init ceilingSqr and squares array
        int next=1, j;
        for(int i=0;i<sqroot;++i){
            squares[i]=next;
            table[next]=1;
            next=(i+2)*(i+2);
            for(j=squares[i];j<next && j<tableLength;++j){
                ceilingSqr[j]=i;
            }
        }
        //find numSquares
        for(int i=2;i<=n;++i){
            if(!table[i])
                recur(i);
        }
        return table[n];
    }
};
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