[LeetCode從零單刷]Perfect Squares

[LeetCode從零單刷]Perfect Squares

題目：

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.

解答：

想一想，當 n = 13 時答案其實是 n = 9 時答案 + 1 的和；n = 12 時答案其實是 n = 8 時答案+ 1 的和，n = 8 時答案其實是 n = 4 時答案 + 1 的和。

這樣，就是明顯的動態規劃了。當 n = i * i 時，返回值為 1；其余情況下，n 依次減去一個小於自身的平方數後求返回值，取所有返回值中最小值再 + 1。

class Solution {
public:
    int numSquares(int n) {
        int max = 1;
        while ((max + 1) * (max + 1) <= n)   max++;
        
        int* dp = new int[n + 1];
        for(int i = 1; i <= max; i++)   dp[i * i] = 1;
        
        int tmp;
        for(int i = 1; i < max; i++) {
            for(int j = i * i + 1; j < (i+1)*(i+1); j++) {
                int min = n;
                for(int k = i; k >= 1; k--) {
                    min = min > (1 + dp[j - k*k]) ? (1 + dp[j - k*k]) : min;
                }
                dp[j] = min;
            }
        }
        
        for(int j = max * max + 1; j <= n; j++) {
            int min = n;
            for(int k = max; k >= 1; k--) {
                min = min > (1 + dp[j - k*k]) ? (1 + dp[j - k*k]) : min;
            }
            dp[j] = min;
        }
        
        return dp[n];
    }
};

但其實，這道題擁有解析解：

根據 Lagrange's four-square theorem，自然數被最少的平方數組合的可能解只會是 1，2，3，4 其中之一。

解 = 1 比較易懂，開方即可；解 = 2, 3, 4 就要依靠 Legendre's three-square theorem 來判斷了。

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 n can be written in the 
        // form of 4^k*(8*m + 7). Please refer to 
        // Legendre's three-square theorem.
        while (n%4 == 0) // n%4 == 0  
            n >>= 2;  

        if (n%8 == 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;  
    }  
};