279. Perfect Squares

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.

这题是一道动态规划题，有点类似于硬币组合成指定值http://blog.csdn.net/u012891472/article/details/51005786
动态规划解法如下：

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