LeetCode 227. Basic Calculator II

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

题目：

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.

题解：

It could apply the generic method. Use two level of operations. Calculate * and / first, accumlate the result into num2.

Then + and -, accumlate the result into num1. 

When current char is digit, get the current number, perform * and /.

If current char is * and /, update operator 2.

If current char is + and -, perform previous + and - operation and update operator 1, reset num2, operator 2.

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

Space: O(1).

AC Java:

class Solution {
    public int calculate(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }
        
        int num1 = 0;
        int o1 = 1;
        int num2 = 1;
        int o2 = 1;
        for(int i = 0; i<s.length(); i++){
            char c = s.charAt(i);
            if(Character.isDigit(c)){
                int cur = c - '0';
                while(i+1 < s.length() && Character.isDigit(s.charAt(i+1))){
                    cur = cur * 10 + s.charAt(i+1)-'0';
                    i++;
                }
                
                num2 = o2 == 1 ? num2 * cur : num2 / cur;
            }else if(c == '*' || c == '/'){
                o2 = c == '*' ? 1 : -1;
            }else if(c == '+' || c == '-'){
                num1 = num1 + o1 * num2;
                o1 = c == '+' ? 1 : -1;
                
                num2 = 1;
                o2 = 1;
            }
        }
        
        return num1 + o1 * num2;
    }
}

类似Basic Calculator, Basic Calculator III.