[lintcode]741. Calculate Maximum Value II-优快云博客

本文介绍了一种算法，通过在给定的数字字符串中插入加号或乘号，并允许使用括号来改变运算优先级，从而找出能获得的最大数值。文章提供了一个具体的实现案例，包括动态规划方法的应用。

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链接：http://www.lintcode.com/zh-cn/problem/calculate-maximum-value-ii/

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.

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.

class Solution {
public:
    /**
     * @param str: a string of numbers
     * @return: the maximum value
     */
    int maxValue(string &str) {
        // write your code here
        int size=str.size();
        if(size==0)
            return 0;
        vector<int> nums;
        for(auto c: str)
            nums.push_back(c-'0');
        vector<vector<int>> dp(size,vector<int>(size,0));
        for(int i=0;i<size;i++)
        {
            dp[i][i]=nums[i];
        }
        for(int step=1;step<size;step++)
        {
            for(int left=0;left<size-step;left++)
            {
                int right=left+step;
                int maxvalue=0;
                for(int mid=left;mid<right;mid++)
                {
                    maxvalue=max(maxvalue,max(dp[left][mid]+dp[mid+1][right],dp[left][mid]*dp[mid+1][right]));
                }
                dp[left][right]=maxvalue;
            }
        }
        return dp[0][size-1];
    }
};