week10- Dynamic Programming-NO.343. Integer Break

题目

• Total Accepted: 39857
• Total Submissions: 87649
• Difficulty: Medium
• Contributor: LeetCode

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.

https://leetcode.com/problems/integer-break/#/description

思路

题目需要将一个整数拆分成若干个数的和，使得这些数的乘积最大

通过观察发现，对于一个大于4的数，要使得乘积最大，需要将其拆分成尽可能多的3的和，最后剩下的数为2或4即可。例如10 = 3 + 3 + 4,14 = 3 + 3 + 3 + 3 + 2。

源程序

class Solution {public:    int integerBreak(int n) {        if(n == 2)            return 1;        if(n == 3)            return 2;        if(n == 4)            return 4;        int r = 1;        while(n > 4){            r *= 3;            n -= 3;        }        r *= n;        return r;    }};