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.

dp[0] = 0 
dp[1] = dp[0]+1 = 1
dp[2] = dp[1]+1 = 2
dp[3] = dp[2]+1 = 3
dp[4] = Min{ dp[4-1*1]+1, dp[4-2*2]+1 } 
      = Min{ dp[3]+1, dp[0]+1 } 
      = 1                               
dp[5] = Min{ dp[5-1*1]+1, dp[5-2*2]+1 } 
      = Min{ dp[4]+1, dp[1]+1 } 
      = 2
dp[13] = Min{ dp[13-1*1]+1, dp[13-2*2]+1, dp[13-3*3]+1 } 
       = Min{ dp[12]+1, dp[9]+1, dp[4]+1 } 
       = 2                                      
dp[n] = Min{ dp[n - i*i] + 1 },  n - i*i >=0 && i >= 1
 

public class Solution {

    public int numSquares(int n) {

       int[] dp = new int[n + 1];

       dp[0] = 0;

       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);   

    //例如：13包括1，4 ，9三個平方數，dp[13-1*1]+1, dp[13-2*2]+1, dp[13-3*3]+1兩兩對比得最小值              

             j++;

          }

          dp[i] = min;

       }

       return dp[n];

   }

}