#Leetcode# 227. Basic Calculator II
阿新 • • 發佈:2018-12-02
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.
程式碼:
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; } };
這樣的四則運算差不多都要用到 $stack$ 吧