[LeetCode]Different Ways to Add Parentheses-优快云博客

本文链接：https://blog.youkuaiyun.com/CiaoLiang/article/details/50997596

本文介绍了一种使用递归算法解决给定包含数字及运算符（加、减、乘）的字符串，计算所有可能组合结果的方法。通过递归分解问题，对每个子表达式求解，最终合并得到所有可能的计算结果。

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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]

用递归来写，符号‘＊’，‘－’，‘＋’，两侧递归求解。

class Solution {
public:
    vector<int> diffWaysToCompute(string str){//以字符串为中心进行递归，对两个子解求解
        vector<int> ret;
        for(int i=0; i<str.size(); ++i){
            if(str[i]=='*'||str[i]=='-'||str[i]=='+'){
                vector<int> left = diffWaysToCompute(str.substr(0,i));
                vector<int> right = diffWaysToCompute(str.substr(i+1));
                for(int j=0; j<left.size(); ++j){
                    for(int k=0; k<right.size(); ++k){
                        if(str[i]=='*')
                            ret.push_back(left[j]*right[k]);
                        else if(str[i]=='+')
                            ret.push_back(left[j]+right[k]);
                        else if(str[i]=='-')
                            ret.push_back(left[j]-right[k]);
                    }
                }
            }
        }
        if(ret.empty())
            ret.push_back(atoi(str.c_str()));
        return ret;
    }
};