PattySung

Given a string s which represents an expression, evaluate this expression and return its value.

The integer division should truncate toward zero.

You may assume that the given expression is always valid. All intermediate results will be in the range of [-231, 231 - 1].

Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().

Example 1:

Input: s = “3+2*2” Output: 7 Example 2:

Input: s = " 3/2 " Output: 1 Example 3:

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

<解題>

注意：是要看前一個存的sign，當前的sign會再更新下一次使用


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

        Stack<Integer> stack = new Stack<>();
        int res = 0;
        int num = 0;
        char sign = '+';

        for (int i = 0; i < s.length(); i++) {
            if (Character.isDigit(s.charAt(i))) {
                num = 0;
                while (i < s.length() && Character.isDigit(s.charAt(i))) {
                    num = num * 10 + (s.charAt(i) - '0');
                    i++;
                }
                i--; // 移除多餘的 i++，因為 for 循環中已經有 i++
            } 

            if(!Character.isDigit(s.charAt(i)) && s.charAt(i) != ' ' || i == s.length()-1){
                if (sign == '+') stack.push(num);
                if (sign == '-') stack.push(-num);
                if (sign == '*') stack.push(stack.pop() * num);
                if (sign == '/') stack.push(stack.pop() / num);
                sign = s.charAt(i);
                num = 0;
            }
        }
        while (!stack.isEmpty()) {
            res += stack.pop();
        }
        return res;
    }
}

Time Complexity : O(n) Space Complexity : O(n)

Leetcode-227. Basic Calculator II

<解題>