Perfect Squares

题目：

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.

解析：

dp[n] indicates that the perfect squares count of the given n, and we have:

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

and the sample code is like below:

class Solution {
public:
    int numSquares(int n) {
        vector<int> dp(n+1,0);
        for(int i = 1; i <= n; ++i)
        {
            int Min = INT32_MAX;
            int j = 1;
            while(i - j * j >= 0)
            {
                Min = min(Min, dp[i-j*j]+1);
                ++j;
            }
            dp[i] = Min;
        }
        return dp[n];
    }
};