Leetcode 224. Basic Calculator

Problem:

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 .

Example 1:

Input: "1 + 1"
Output: 2

Example 2:

Input: " 2-1 + 2 "
Output: 3

Example 3:

Input: "(1+(4+5+2)-3)+(6+8)"
Output: 23

Note:

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

 

Solution:

　　两种解法，递归解法碰到左括号时，找到与之匹配的右括号，将括号内的字符串调用递归函数进行计算，得到最后结果为止，递归解法不需要用栈，但是由于提取子字符串的操作导致效率低下。

class Solution {
public:
    int calculate(string s) {
        int pointer = 0;
        int number = 0;
        int sign = 1;
        int result = 0;
        while(pointer != s.size()){
            if(s[pointer] >= '0' && s[pointer] <= '9'){
                number = number*10+s[pointer]-'0';
            }
            else if(s[pointer] == '('){
                int left = pointer;
                int right = pointer;
                int count = 1;
                while(count != 0){
                    right++;
                    if(s[right] == '(')
                        count++;
                    else if(s[right] == ')')
                        count--;
                }
                result += sign*calculate(s.substr(left+1,right-left-1));
                pointer = right;
            }
            else if(s[pointer] == '-' || s[pointer] == '+' || pointer == s.size()-1){
                result += sign*number;
                number = 0;
                sign = (s[pointer] == '+')?1:-1;
            }
            pointer++;
        }
        result += sign*number;
        return result;
    }
};

　　第二种用栈的解法，当碰到左括号时，栈中压入之前得到的结果以及符号，当碰到右括号时，将括号内的结果乘以栈顶符号再加上符号左侧的结果，效率高。

Code:

 

class Solution {
public:
    int calculate(string s) {
        stack<int> stk;
        int pointer = 0;
        int number = 0;
        int sign = 1;
        int result = 0;
        while(pointer != s.size()){
            if(s[pointer] >= '0' && s[pointer] <= '9'){
                number = number*10+s[pointer]-'0';
            }
            else if(s[pointer] == '('){
                stk.push(result);
                stk.push(sign);
                result = 0;
                sign = 1;
            }
            else if(s[pointer] == ')'){
                result += sign*number;
                number = 0;
                result *= stk.top();
                stk.pop();
                result += stk.top();
                stk.pop();
            }
            else if(s[pointer] == '-' || s[pointer] == '+' || pointer == s.size()-1){
                result += sign*number;
                number = 0;
                sign = (s[pointer] == '+')?1:-1;
            }
            pointer++;
        }
        result += sign*number;
        return result;
    }
};