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;
}

转载于:https://www.cnblogs.com/chess/p/5296502.html

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