动态规划-343. Integer Break

题目：
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.

假设N为4

image.png

先使用记忆化搜索的自顶向下的递归解法：

public class P343IntegerBreak {

    public int integerBreak(int n) {

        return digui(n);
    }
    // 求解的是分解n的最优乘积.
    private int digui(int n) {
        if (n == 1) {
            return 1;
        }
        int max = -1;
        for (int i = 1; i < n; i++) {
            max = Math.max(max, i * digui(n - i));
            max = Math.max(max, i * (n - i));
        }
        return max;
    }
    public static void main(String[] args) {

        int n = 10;
        P343IntegerBreak p = new P343IntegerBreak();
        int i = p.integerBreak(n);
        System.out.println(i);
    }

}

从图里面其实可以看到比如分解2最大的乘积实际上会被多计算几次的，因此实际上可以记录下来他的值，防止重复计算。开辟一个大小为n的数组进行保存。代码如下：

public class P343IntegerBreakV2 {

    int[] tmp;
    public int integerBreak(int n) {
        tmp = new int[n+1];
        for (int i = 0; i < n+1; i++) {
            tmp[i] = -1;
        }
        return digui(n);
    }
    // 求解的是分解n的最优乘积.
    private int digui(int n) {
        if (n == 1) {
            return 1;
        }
        if (tmp[n] != -1) {
            return tmp[n];
        }
        int max = -1;
        for (int i = 1; i < n; i++) {
            max = Math.max(max, i * digui(n - i));
            max = Math.max(max, i * (n - i));
        }
        tmp[n] = max;
        return max;
    }
    public static void main(String[] args) {

        int n = 2;
        P343IntegerBreakV2 p = new P343IntegerBreakV2();
        long start = System.currentTimeMillis();
        int i = p.integerBreak(n);
        System.out.println(i);
        System.out.println("cost time = " + (System.currentTimeMillis() - start));
    }

}

如果使用自底向上的动态规划算法的话，代码如下：

public class P343IntegerBreakV3 {

    // 存储分解数字n(分解成2部分）的最大乘积和。
    int[] tmp;
    public int integerBreak(int n) {
        tmp = new int[n+1];
        for (int i = 0; i < n+1; i++) {
            tmp[i] = -1;
        }
        tmp[1] = 1;
        for (int i = 1; i <= n; i++) {
            // 将数字i 分解为 j + (i-j)
            for (int j = 1; j < i; j++) {
                tmp[i] = Math.max(j * (i - j),Math.max(tmp[i], j * tmp[i-j]));
            }
        }
        return tmp[n];
    }
    public static void main(String[] args) {

        int n = 10;
        P343IntegerBreakV3 p = new P343IntegerBreakV3();
        long start = System.currentTimeMillis();
        int i = p.integerBreak(n);
        System.out.println(i);
        System.out.println("cost time = " + (System.currentTimeMillis() - start));
    }

}