leetcode241 : Different Ways to Add Parentheses-优快云博客

本文链接：https://blog.youkuaiyun.com/l3368bcttqnqn/article/details/49508933

本文介绍了一种通过递归算法解决给定包含数字及运算符的字符串，计算所有可能组合结果的方法。针对不同运算符，算法实现了有效的递归拆分与计算。

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1、原题如下：
Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *.

Example 1
Input: “2-1-1”.

((2-1)-1) = 0
(2-(1-1)) = 2
Output: [0, 2]

Example 2
Input: “2*3-4*5”

(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10
Output: [-34, -14, -10, -10, 10]

2、解题如下：

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> result;
        int offset=0;//用来弥补string中没有+-*的情况
        for(int i=0;i<input.size();i++)
        {
            if(input[i]=='+'||input[i]=='-'||input[i]=='*')
            {
                offset=1;
                vector<int> temp1=diffWaysToCompute(input.substr(0,i));
                vector<int> temp2=diffWaysToCompute(input.substr(i+1));
                for(auto x:temp1)
                {
                    for(auto y:temp2)
                    {
                        if(input[i]=='+')
                            result.push_back(x+y);
                        if(input[i]=='-')
                            result.push_back(x-y);
                        if(input[i]=='*')
                            result.push_back(x*y);
                    }
                }
            }
        }
        if(offset==0)
        {
            result.push_back(stoi(input));
        }
        return result;
    }
};