LeetCode-343.Integer Break

https://leetcode.com/problems/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.

Hint:
There is a simple O(n) solution to this problem.
You may check the breaking results of n ranging from 7 to 10 to discover the regularities.
方法1
多列举几个数就可以发现规律

int integerBreak(int n)
{
if (n <= 3)
return n-1;
if (n % 3 == 0)
return pow(3, n / 3);
if (n % 3 == 1)
return pow(3, n / 3 - 1) * 4;
return pow(3, n / 3) * 2;
}

方法2
动态规划 一个数要么拆分成2，要么拆分成3

int integerBreak(int n)
{
if (n <= 3)
return n-1;
vector<int> dp(n - 3);
dp[0] = 4;//n=4
dp[1] = 6;//n=5
dp[2] = 9;//n=6
for (int i = 3; i < n - 3; i++)
dp[i] = max(3 * dp[i - 3], 2 * dp[i - 2]);
return dp[n-4];
}