leetcode@ [343] Integer Break (Math & Dynamic Programming)

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:

Show Hint

  1. There is a simple O(n) solution to this problem.Show More Hint
  2. You may check the breaking results of n ranging from 7 to 10 to discover the regularities.
public class Solution {
    public int integerBreak(int n) {
        
        if(n <= 2) {
            return 1;
        } else if (n == 3) {
            return 2;
        }
        
        double radical = Math.sqrt((double) n);
        int rad = (int) Math.floor(radical);
        
        int rs = 0, power = 0;
        for(int i=1; i<=rad+1; ++i) {
            int times = n / i;
            int remain = n - i*times;
            
            if(remain == 0) {
                power = (int) Math.pow(i, times);
                rs = Math.max(rs, power);
            } else {
                int power_1 = (int) Math.pow(i, times-1) * (remain + i);
                int power_2 = (int) Math.pow(i, times) * remain;
                rs = Math.max(rs, Math.max(power_1, power_2));
            }
        }
        
        return rs;
    }
}
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原文地址:https://www.cnblogs.com/fu11211129/p/5628645.html