LeetCode 224. Basic Calculator

原题链接在这里：https://leetcode.com/problems/basic-calculator/

题目：

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 

题解：

The final resut is num1 + o1 * num2.

First remvoe all the whitespaces.

 

遇到数字位，看后一位是否为数字，若是位数字，当前位需要进十. Update num2.

For operator, if it is the first operator or right after "(", just update o1. Otherwise, num1 = num1 + o1 * num2. Then update o1 and reset o1 and num2.

For "(", push into stack. For ")", cur = num1 + o1 * num2. pop the stack and num2 = cur.

Time Complexity: O(s.length()).

Space: O(s.length()), stack space.

AC Java: 

class Solution {
    public int calculate(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }

        s = s.replaceAll("\\s+", "");
        int num1 = 0;
        int o1 = 1;
        int num2 = 1;
        Stack<Integer> stk = new Stack<>();
        int n = s.length();
        for(int i = 0; i < n; i++){
            char c = s.charAt(i);
            if(Character.isDigit(c)){
                int cur = (int)(c - '0');
                while(i + 1 < n && Character.isDigit(s.charAt(i + 1))){
                    cur = cur * 10 + (int)(s.charAt(i + 1) - '0');
                    i++;
                }

                num2 = cur;
            }else if(c == '+' || c == '-'){
                if(c == '-' && (i == 0 || s.charAt(i-1) == '(')){
                    o1 = -1;
                    continue;
                }

                num1 = num1 + o1 * num2;
                o1 = c == '+' ? 1 : -1;
                num2 = 1;
            }else if(c == '('){
                stk.push(num1);
                stk.push(o1);
                stk.push(num2);

                num1 = 0;
                o1 = 1;
                num2 = 1;
            }else if(c == ')'){
                int cur = num1 + o1 * num2;
                num2 = stk.pop();
                o1 = stk.pop();
                num1 = stk.pop();
                num2 = cur;
            }
        }

        return num1 + o1 * num2;
    }
}

 

类似Evaluate Reverse Polish Notation.

跟上Basic Calculator II.