Given a string of numbers, write a function to find the maximum value from the string, you can add a + or * sign between any two numbers，It’s different with Calculate Maximum Value, You can insert parentheses anywhere.

樣例
Given str = 01231, return 12
(0 + 1 + 2) * (3 + 1) = 12 we get the maximum value 12

Given str = 891, return 80
As 8 * (9 + 1) = 80, so 80 is maximum.

思路：
構建陣列res[][]，res[i][j]表示從str[i]到str[j]的字串，可以取得的最大值。
在不同步長的情況下，計算不同位置的值：對於01231
如先計算[0,1],[1,2],[2,3]… 再計算[0,1,2],[1,2,3]…
遞迴求解在位置為i，步長為path的情況下，由字串可以取得的最大值，並將其賦值給res[i][i+path-1]。

#ifndef C741_H
#define C741_H
#include<iostream>
 

#include<string>
#include<vector>
using namespace std;
class Solution {
public:
    /**
    * @param str: a string of numbers
    * @return: the maximum value
    */
    int maxValue(string &str) {
        // write your code here
        if (str.empty())
            return 0;
        int
 
 len = str.size();
        vector<vector<int>> res(len, vector<int>(len, INT_MIN));//res[i][j]存放從str[i]到str[j]的最大值
        for (int i = 0; i<len; ++i)//初始化res
            res[i][i] = str[i] - '0';
        //在不同步長不同位置的情況下計算res的值
        for (int path = 2; path <= len; ++path)
        {
            for (int i = 0; i + path <= len;++i)
                res[i][i + path - 1] = ComputeMax(i, path, res);
        }
        return res[0][len - 1];
    }
    //遞迴計算位置i,步長為path的情況下取得的最大值
    int ComputeMax(int i,int path,vector<vector<int>> vec) {
        int res = INT_MIN;
        if (vec[i][i + path - 1] != INT_MIN)
            return vec[i][i + path - 1];
        for (int k = 1; k < path; ++k) {
                int left = ComputeMax(i, k, vec);
                int right = ComputeMax(i + k, path - k, vec);
                res = maxVal(maxVal(left + right, left*right),res);
            }
        return res;
    }
    int maxVal(int a, int b)
    {
        return a > b ? a : b;
    }
};
#endif