[LeetCode] | Basic Calculator II

https://leetcode.com/problems/basic-calculator-ii/?tab=Solutions

思路1：Stack

开一个栈放整数，遇到加减号就压栈，遇到乘除号就取出栈顶与当前数做乘除再压回去。

基础操作：O(n)去掉字符串中的所有空格，取字符串的下一个整数。

Time complexity: O(n)
Space complexity: O(n)

class Solution {
public:
    int calculate(string s) {
        if (s.empty()) return 0;
        
        // remove all the blanks in the string
        s.erase(remove(s.begin(), s.end(), ' '), s.end());

        stack<int> stk;
        string str = get_next(s, 0);
        stk.push(stoi(str));
        
        int i = str.size();
        while (i < s.size()) {
            char oper = s[i];
            string str = get_next(s, i + 1);  // get the current number
            int cur_num = stoi(str);
            
            if (oper == '+') {
                stk.push(cur_num);
            } else if (oper == '-') {
                stk.push(-cur_num);
            } else if (oper == '*') {
                int top = stk.top(); stk.pop();
                stk.push(top * cur_num);
            } else {
                int top = stk.top(); stk.pop();
                stk.push(top / cur_num);
            }
            
            i += (1 + str.size());
        }
        
        int ret = 0;
        while (!stk.empty()) {
            ret += stk.top();
            stk.pop();
        }
        
        return ret;
    }
    
private:
    string get_next(string& s, int pos) {
        int end = pos;
        while (end < s.size() && isdigit(s[end])) {
            end++;
        }
        return s.substr(pos, end - pos);
    }
};

思路2：用trick实现O(1)的space

/*
 * author  : TK
 * date    : 2017-02-22
 * problem : LeetCode 227. Basic Calculator
 * tags    : String
 * language: C++
 */
// key points: x+y*z = [(x+y)-y]+y*z, which make it possible to realize O(1) space
// Time complexity: O(n)
// Space complexity: O(1)
class Solution {
public:
    int calculate(string s) {
        if (s.empty()) return 0;
        
        // remove all the blanks in the string
        s.erase(remove(s.begin(), s.end(), ' '), s.end());
        
        string str = get_next(s, 0);  // get the first number
        int prev = stoi(str), ret = prev;
        
        int i = str.size();
        while (i < s.size()) {
            char oper = s[i];
            string str = get_next(s, i + 1);  // get the current number
            int cur_num = stoi(str);
            
            if (oper == '+') {
                ret += cur_num;
                prev = cur_num;
            } else if (oper == '-') {
                ret -= cur_num;
                prev = -cur_num;
            } else if (oper == '*') {
                ret = (ret - prev) + prev * cur_num;
                prev *= cur_num;
            } else {
                ret = (ret - prev) + prev / cur_num;
                prev /= cur_num;
            }
            
            i += (1 + str.size());
        }
        
        return ret;
    }
    
private:
    string get_next(string& s, int pos) {
        int end = pos;
        while (end < s.size() && isdigit(s[end])) {
            end++;
        }
        return s.substr(pos, end - pos);
    }
};

提醒：写完以后提交却妥妥地超时，检查后发现get_next()函数中的参数s没加引用，这可就灾难了，因为不加引用会进行字符串的整体copy...