leetcode 241: Different Ways to Add Parentheses

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]

思路：分治，将exp根据运算符拆分成左右两个子表达式，分别计算，合并；（一个表达式遍历运算符，拆分多次）

注意：input[i]-'0'

注意：结果集不必去重

class Solution {
public:
    vector<int> diffWaysToCompute(string input) {
        vector<int> ans;
        int len = input.length();
        int flag=0;
        int i;
        for(i=0;i<len;i++)
        {
            char c = input[i];
            if(c=='+'||c=='-'||c=='*')
            {
                flag=1;
                vector<int> left = diffWaysToCompute(input.substr(0,i));
                vector<int> right = diffWaysToCompute(input.substr(i+1,len-i-1));
                for(int j=0;j<left.size();j++)
                {
                    for(int k=0;k<right.size();k++)
                    {
                        int tmp=0;
                        if(c == '+')
                            tmp = left[j]+right[k];
                        if(c == '-')
                            tmp = left[j]-right[k];
                        if(c == '*')
                            tmp = left[j]*right[k];
                        ans.push_back(tmp);
                    }
                }
            }
        }
        if(flag==0)
        {
            int tmp=0;
            for(i=0;i<len;i++)
            {
                tmp = tmp*10 + input[i]-'0';
            }
            ans.push_back(tmp);
        }
        //vector<int> ans2(ans.begin(),ans.end());
        return ans;
    }
};

 附：

set<int> s;
s.insert(1);
set<int> s(v.begin(),v.end());
vector<int> v(s.begin(),s.end());