241. Different Ways to Add Parentheses

241. Different Ways to Add Parentheses https://leetcode.com/problems/different-ways-to-add-parentheses/

思路就是：首先找到以运算符为根节点，分别计算左子串和右子串的所有结果的集合，然后依次进行组合计算。参考博客/article/5095878.html。

自己的思路错了，直接用两边只用了一个整数去接收左右子串的计算值！！

#include<iostream>
#include<vector>
#include<string>
using namespace std;
class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> temp;
for (int i = 0; i < input.size(); i++){
if (isop(input[i])){
vector<int> left = diffWaysToCompute(input.substr(0, i));
vector<int> right = diffWaysToCompute(input.substr(i + 1));
for (int j = 0; j < left.size(); j++){
for (int k = 0; k < right.size(); k++){
temp.push_back(compute(left[j], right[k], input[i]));
}
}
}
}
if (temp.empty()){
temp.push_back(atoi(input.c_str()));
}
return temp;
}

bool isop(char ch){
if (ch == '+' || ch == '-' || ch == '*')
return true;
return false;
}

int compute(int v1, int v2, char ch){
int sum = 0;
switch (ch){
case '+':
sum = v1 + v2; break;
case '-':
sum = v1 - v2; break;
case '*':
sum = v1*v2; break;
}
return sum;
}

};
int main()
{
Solution test;
string te = "2+4*3";
vector<int> res=test.diffWaysToCompute(te);
for (auto it = res.begin(); it != res.end(); it++){
cout << *it << endl;
}

return 0;
}