题目描述

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.

Example 1:

Input: 2
Output: 1
Explanation: 2 = 1 + 1, 1 × 1 = 1.

Example 2:

Input: 10
Output: 36
Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.

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

解题思路

该题就是 剪绳子问题的数字描述 版本。从上到下思考问题，自底向上解决问题

计算N=10

那么需要计算 1+9 2+8 3+7 4+6 5+5

为什么只需要考虑切一次呢？

因为这里的 9 8 7 6 5 都已经是经过最好切分的结果了

10 = 1+ 9 = 1+ 3+3+3 1x3x3x3 = 27 。。。

当计算出前面问题的最优解 即可

题解

public int integerBreak(int n) {
    // not less than 2
    if(n == 2) return 1;
    if(n == 3) return 2;
    int mem[] = new int[n+1];
    // 这里从4开始 才是最优解的值，前面的存储都是为了乘法预先保存的
    mem[0] = 0;
    mem[1] = 1;
    mem[2] = 2;
    mem[3] = 3;

    for (int i = 4; i <= n; i++) {
        int max = 0;
        for (int j = 1; j <= i/2; j++) {
            int prod = mem[j]*mem[i-j];
            if(prod > max) max = prod;
        }
        mem[i] = max;
    }
    
    return mem[n];
}