[LintCode] Perfect Squares 完全平方数

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.

Example
Given n = 12, return 3 because 12 = 4 + 4 + 4
Given n = 13, return 2 because 13 = 4 + 9

LeetCode上的原题，请参见我之前的博客Perfect Squares。

解法一：

class Solution {
public:
    /**
     * @param n a positive integer
     * @return an integer
     */
    int numSquares(int n) {
        while (n % 4 == 0) n /= 4;
        if (n % 8 == 7) return 4;
        for (int a = 0; a * a <= n; ++a) {
            int b = sqrt(n - a * a);
            if (a * a + b * b == n) {
                return !!a + !!b;
            }
        }
        return 3;
    }
};

解法二：

class Solution {
public:
    /**
     * @param n a positive integer
     * @return an integer
     */
    int numSquares(int n) {
        while (n % 4 == 0) n /= 4;
        if (n % 8 == 7) return 4;
        vector<int> dp(n + 1, INT_MAX);
        dp[0] = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 1; i + j * j <= n; ++j) {
                dp[i + j * j] = min(dp[i + j * j], dp[i] + 1);
            }
        }
        return dp.back();
    }
};

解法三：

class Solution {
public:
    /**
     * @param n a positive integer
     * @return an integer
     */
    int numSquares(int n) {
        while (n > 0 && n % 4 == 0) n /= 4;
        if (n % 8 == 7) return 4;
        int res = n, i = 2;
        while (i * i <= n) {
            int a = n / (i * i), b = n % (i * i);
            res = min(res, a + numSquares(b));
            ++i;
        }
        return res;
    }
};