Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.

Example 1:

Input: n = 12
Output: 3 
Explanation: 12 = 4 + 4 + 4.
Example 2:
Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.

方法1：動態規劃

 int[] dp = new int[n+1];
        dp[0] = 0;
        for (int i = 1; i <= n; i++) {
            int min = Integer.MAX_VALUE;
            for (int k = 1; k * k <= i; k++) {
                min = Math.min(min, 1+dp[i-k*k]);
            }
            dp[i] = min;
        }
        return dp[n];
 

時間複雜度:O(n^2)

空間複雜度:O(n)