問題

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.

You may assume that the given expression is always valid.

例子

"3+2*2" = 7
" 3/2 " = 1
" 3+5 / 2 " = 5

分析

1 使用istringstream，將字符串變成字符流，自動(dòng)分割字符串并直接轉(zhuǎn)換成數(shù)字或字符
2 使用棧，保存表達(dá)式中的數(shù)字或者兩個(gè)數(shù)字*/的結(jié)果
3 遍歷棧，將棧元素依次相加即可得到表達(dá)式的計(jì)算結(jié)果

舉個(gè)例子，對(duì)于表達(dá)式3-5/2，首先初始化運(yùn)算符為+號(hào)，即表達(dá)式變成+3-5/2；將+3壓入棧，-5壓入棧，發(fā)現(xiàn)下一個(gè)運(yùn)算符是/，將-5出棧，將-5/2=-2入棧。

要點(diǎn)

• istringstream可以處理任意多空格分割的字符串
• 用棧來保存上一步的數(shù)字
• 用vector模擬棧

時(shí)間復(fù)雜度

O(n)

空間復(fù)雜度

O(n)

代碼

class Solution {
public:
    int calculate(string s) {
        vector<int> stk;
        istringstream iss(s);
        char op = '+';
        int res = 0, term;
        while (iss >> term) {
            if (op == '+' || op == '-') {
                stk.push_back(op == '+' ? term : -term);
            }
            else {
                int last = stk.back();
                stk.pop_back();
                stk.push_back(op == '*' ? last * term : last / term);
            }
            iss >> op;
        }
        for (int num : stk)
            res += num;
        return res;
    }
};