华为的一道上机题，一个简单的计算算术表达式（带括号）的程序，没有使用传统的栈来解决。

expression.cpp

#include <iostream>
#include <string>
#include <vector>
#include <sstream>

#include <cstdlib>

using namespace std;

static int simple_arithmetic(const string &k_expression_) // No any brakets.
{
    vector<int> nums;
    vector<char> todo;  // for all operational character.
    vector<int>::iterator p_nums;
    size_t len = k_expression_.size();
    int tmp;
    size_t i = 0, j = 0;

    for (; i < len; ++i) {
        if (k_expression_[i] == '*' || k_expression_[i] == '/' ||   \
                k_expression_[i] == '-' || k_expression_[i] == '+') { 
            if (i != j) {
                nums.push_back(atoi(k_expression_.substr(j, i - j).c_str()));
                todo.push_back(k_expression_[i]);
                j = i + 1;
                continue;
            }
            else { // negative number.
                nums.push_back(0 - atoi(k_expression_.substr(1).c_str()));
                break;
            }
        }
    }
    nums.push_back(atoi(k_expression_.substr(j, i - j).c_str())); // to add the last numbers.

    for (vector<char>::iterator i = todo.begin(); i != todo.end();) {
        if (*i == '*') {
            tmp = nums[i - todo.begin() + 1];
            nums.erase(nums.begin() + (i - todo.begin()) + 1);
            nums[i - todo.begin()] *= tmp;
            i = todo.erase(i); // I can't erase current vector's element when I'm in it.
        }
        else if (*i == '/') {
            tmp = nums[i - todo.begin() + 1];
            nums.erase(nums.begin() + (i - todo.begin()) + 1);
            nums[i - todo.begin()] /= tmp;
            i = todo.erase(i);
        }
        else
            i++;
    }

    for (vector<char>::iterator i = todo.begin(); i != todo.end();) {
        if (*i == '+') {
            tmp = nums[i - todo.begin() + 1];
            nums.erase(nums.begin() + (i - todo.begin()) + 1);
            nums[i - todo.begin()] += tmp;
            i = todo.erase(i);
        }
        else if (*i == '/') {
            tmp = nums[i - todo.begin() + 1];
            nums.erase(nums.begin() + (i - todo.begin()) + 1);
            nums[i - todo.begin()] -= tmp;
            i = todo.erase(i);
        }
        else
            i++;
    }

    return *nums.begin();
}

int parentheses_arithmetic(const string &k_expression_)
{
    string expression(k_expression_);
    string tmp("");
    stringstream ss;
    int rpos, lpos = 0;

    rpos = expression.find(')');
    while (rpos != string::npos) {
        lpos = expression.rfind('(', rpos);
        ss << simple_arithmetic(expression.substr(lpos + 1, rpos - lpos - 1));
        tmp = ss.str();
        ss.str(""); // clear the stringstream.
        expression.replace(lpos, rpos - lpos + 1, tmp);
        rpos = expression.find(')');
    }

    return simple_arithmetic(expression);
}

int main()
{
    string ex = "(1+2+(3+4))+(5/6)";
    cout << parentheses_arithmetic(ex) << endl;
}