[Leetcode] #343 Integer Break (DP)

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本文介绍了一种算法，用于将一个正整数拆分成至少两个正整数之和，并最大化这些整数的乘积。通过分析发现，当拆分后的整数接近3时，乘积最大。提供了两种解决方案：一种是直观的迭代法，另一种是使用动态规划。

Discription

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.

Solution

class Solution {
public:
	// As the hint said, checking the n with ranging from 7 to 10 to discover the regularities.
	// n = 7,    3*4 = 12
	// n = 8,  3*3*2 = 18
	// n = 9,  3*3*3 = 27
	// n = 10, 3*3*4 = 36
	// n = 11, 3*3*3*2 = 54
	//
	// we can see we can break the number by 3 if it is greater than 4;
	//
	int integerBreak(int n) {
		if (n == 2) return 1;
		if (n == 3) return 2;
		int result = 1;
		while (n > 4) {
			result *= 3;
			n -= 3;
		}
		result *= n;
		return result;
	}

	int integerBreak1(int n) {
		vector<int> dp(n + 1, 1);
		for (int i = 2; i <= n; i++) {
			for (int j = 1; j <= i - 1; j++) {
				if (i - j <= 4)
					dp[i] = max((i - j)*j, dp[i]);
				else
					dp[i] = max(dp[i - j] * j, dp[i]);
			}
		}
		return dp[n];
	}
};

GitHub-Leetcode: https://github.com/wenwu313/LeetCode