leetcode: Basic Calculator
Implement a basic calculator to evaluate a simple expression string. The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces . You may assume that the given expression is always valid. Some examples:
“1 + 1” = 2 ” 2-1 + 2 ” = 3 “(1+(4+5+2)-3)+(6+8)” = 23
/*Example: 7 – (6 – 5 – (4 – 3) – 2) – (1)Result : +7 – 6 + 5 + 4 – 3 + 2 – 1The + -are signs just before ‘(‘.They are dealt with previous stack sign.Then they are saved in the stack to help decide the signs in the following “(..)” .*/class Solution {public:int calculate(string s) {stack<int> t; //previous sign just before the ‘(‘t.push(1); //for all the signs those are not in the parenthesesint sum = 0, temp = 0, lastSign = 1;for (auto c : s){(c == ‘)’) t.pop();if (c >= ‘0’ && c <= ‘9’){temp = temp * 10 + c – ‘0’;}if (c == ‘-‘ || c == ‘+’){sum += lastSign*temp; //calculate the number before current signtemp = 0;lastSign = (c == ‘-‘ ? -1 : 1)*t.top(); //deal with the stack influencing sign}}sum += lastSign*temp; //calculate the last numberreturn sum;}};
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