241. Different Ways to Add Parentheses

Posted 2020-06-25 棋子

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http://www.cnblogs.com/ganganloveu/p/4681439.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;
}