279. Perfect Squares

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本文探讨了如何找出构成正整数n所需的最少完全平方数数量，通过动态规划算法解决这一经典问题，并提供了一个C++实现示例。

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.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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参考discusshttps://leetcode.com/discuss/58056/summary-of-different-solutions-bfs-static-and-mathematics

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. Note that cntPerfectSquares[0] is 0.
        vector<int> cntPerfectSqures(n+1,INT_MAX);
        cntPerfectSqures[0] = 0;
        for(int i = 1;i<n+1;i++)
        {
            // For each i, it must be the sum of some number (i - j*j) and 
                // a perfect square number (j*j).
            for(int j = 1;j*j<=i;j++)
            {
                
                cntPerfectSqures[i] = min(cntPerfectSqures[i],cntPerfectSqures[i-j*j]+1);
            }
        }
        
        return cntPerfectSqures.back();
    }
};