#Leetcode# 227. Basic Calculator II

https://leetcode.com/problems/basic-calculator-ii/

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.

Example 1:

Input: "3+2*2"
Output: 7

Example 2:

Input: " 3/2 "
Output: 1

Example 3:

Input: " 3+5 / 2 "
Output: 5

Note:

  • You may assume that the given expression is always valid.
  • Do not use the eval built-in library function.

代码:

class Solution {
public:
    int calculate(string s) {
        int ans = 0, num = 0, n = s.size();
        char op = '+';
        stack<int> st;
        for (int i = 0; i < n; ++i) {
            if (s[i] >= '0' && s[i] <= '9') {
                num = num * 10 + s[i] - '0';
            }
            if ((s[i] < '0' && s[i] != ' ') || i == n - 1) {
                if (op == '+') st.push(num);
                if (op == '-') st.push(-num);
                if (op == '*' || op == '/') {
                    int tmp = (op == '*') ? st.top() * num : st.top() / num;
                    st.pop();
                    st.push(tmp);
                }
                op = s[i];
                num = 0;
            } 
        }
        while (!st.empty()) {
            ans += st.top();
            st.pop();
        }
        return ans;
    }
};

  这样的四则运算差不多都要用到 $stack$ 吧

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转载自www.cnblogs.com/zlrrrr/p/10054362.html