279. Perfect Squares

题意： 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.

思路：这题和之前322.Coins Change差不多，只是硬币变成了平方数，具体思路还是DP做，递推式都和之前那题一样：d[i]=min(d[i],d[i−c[i]]+1)，每个数都由平方数组合而成，具体见代码：

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

注意到其实在对应计算当前i时，之前的已经计算完了，有种递归的思路在里面，而且这题和322的题不同，它相当于隐含了每个数都有解的意思在里面。但是这个方法还有一个问题，就是问题题目判定程序都要重新构造一个长为n+1的数组，很费时间，所以Solutions里就有人写了下面这个新的方法：

class Solution {public:    int numSquares(int n)     {        if (n <= 0)        {            return 0;        }        // cntPerfectSquares[i] = the least number of perfect square numbers         // which sum to i. Since cntPerfectSquares is a static vector, if         // cntPerfectSquares.size() > n, we have already calculated the result         // during previous function calls and we can just return the result now.        static vector<int> cntPerfectSquares({0});        // While cntPerfectSquares.size() <= n, we need to incrementally         // calculate the next result until we get the result for n.        while (cntPerfectSquares.size() <= n)        {            int m = cntPerfectSquares.size();            int cntSquares = INT_MAX;            for (int i = 1; i*i <= m; i++)            {                cntSquares = min(cntSquares, cntPerfectSquares[m - i*i] + 1);            }            cntPerfectSquares.push_back(cntSquares);        }        return cntPerfectSquares[n];    }};

这个静态方法中vector的声明只会被执行一次，所以之后小于上次执行函数中n的解，就可以直接从静态数组中读取出来了，当然这题还有一个数学的方法，也就是我之前说的这题必有解的原因，Lagrange’s four-square theorem，简单来说，就是一个自然数，必定能拆分成四个平方数（当然这里0也算平方数），所以只要求解出解是1~4之中哪个，再结合Legendre’s three-square theorem 就可以找出解是哪个了，具体看参考代码：

class Solution {  private:      int is_square(int n)    {          int sqrt_n = (int)(sqrt(n));          return (sqrt_n*sqrt_n == n);      }public:    // Based on Lagrange's Four Square theorem, there     // are only 4 possible results: 1, 2, 3, 4.    int numSquares(int n)     {          // If n is a perfect square, return 1.        if(is_square(n))         {            return 1;          }        // The result is 4 if and only if n can be written in the         // form of 4^k*(8*m + 7). Please refer to         // Legendre's three-square theorem.        while ((n & 3) == 0) // n%4 == 0          {            n >>= 2;          }        if ((n & 7) == 7) // n%8 == 7        {            return 4;        }        // Check whether 2 is the result.        int sqrt_n = (int)(sqrt(n));         for(int i = 1; i <= sqrt_n; i++)        {              if (is_square(n - i*i))             {                return 2;              }        }          return 3;      }  }; 

最后还有一个BFS的python的代码，比较有意思，也放上来

def numSquares(self, n):    if n < 2:        return n    lst = []    i = 1    while i * i <= n:        lst.append( i * i )        i += 1    cnt = 0    toCheck = {n}    while toCheck:        cnt += 1        temp = set()        for x in toCheck:            for y in lst:                if x == y:                    return cnt                if x < y:                    break                temp.add(x-y)        toCheck = temp    return cnt