leetcode 343. Integer Break

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探讨如何通过动态规划解决整数拆分问题，即给定一个正整数n，将其拆分为至少两个正整数之和，并最大化这些整数的乘积。

343. Integer Break

Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.

For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4).

Note: You may assume that n is not less than 2 and not larger than 58.

1、动态规划的思想。

2、中间有一步。如果拆分后的乘积还没有本身大，那就不拆分。

class Solution {
public:
    int integerBreak(int n)
    {
        if (n == 2) return 1;

        vector<int> ret(n + 1, 0);
        ret[1] = 1;
        ret[2] = 2;
        for (int i = 3; i <= n; i++)
        {
            for (int j = 1; j <= i/2; j++ )
            {
                ret[i] = max(ret[i], ret[j] * ret[i - j]);
            }
            if (i < n) ret[i] = max(ret[i], i);
        }
        return ret[n];
    }
};