Perfect Squares——动态规划

问题描述

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.

算法分析

考虑Sum = a + b + c + d，a,b,c,d均是perfect square number，用F[Sum]表示最小numbers，即F[Sum] =3。

1、对于b + c + d，F[Sum-a] + 1 = F[Sum]，同样F[c+d] + 1 = F[b+c+d] ，F[d] + 1 = F[c+d]。

2、从1可以看出这就是动态规划的思想，定义状态F[N] ，状态转移方程为：F[N] = min(F[Sum-i^2]+1) ( i=1,2,3...N^(1/2) ) 。

3、返回的F[N]即为the least number of perfect square numbers。

​

C++实现

int numSquares(int n) {
        if (n <= 0) return 0;

        // cntPerfectSquares[i]表示和为i的Perfect Squares. 从i=1开始计算，动态规划的思想计算所有子解
        static vector<int> cntPerfectSquares({0});

        // 计算cntPerfectSquares[i]知道i = n，即cntPerfectSquares.size() == n+1
        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];
    }