Leetcode[227] Basic Calculator II

LeetCode[227] Basic Calculator II

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.

You may assume that the given expression is always valid.

Some examples:

"3+2*2" = 7
" 3/2 " = 1
" 3+5 / 2 " = 5

Stack

复杂度
O(N), O(N)

思路
用两个stack来分别记录当前的结果和操作符(operator), 注意每一次统计当前的operand的时候，要看一下下一位的操作符。如果下一位是high operator('+'或'/'), 就要先把当前的数和操作符都要先push回stack保留数据，如果当前stack不为空且已经有高位运算符在stack的顶部，就先计算一次重新计算当前的数字，再考虑之后的运算。

有一种replace all multiple spaces with one space的方法，是s.replaceAll("\\s+",""), \s 表示的是匹配任意的空白符，包括空格，制表符(Tab)，换行符，中文全角空格等。也可以用更简单的方法，replaceAll(" +", "")。

代码

public double calculator(String s) {
    // delete multiple spaces
    s = s.trim().replaceAll(" +", "");
    char[] arr = s.toCharArray();
    Stack<Double> res = new Stack<>();
    Stack<Character> operator = new Stack<>();
    res.push(0.0);
    operator.push('+');
    for(int i = 0; i < arr.length; i ++) {
        double val = 0;
        int j = i;
        while(j < arr.length && Character.isDigit(arr[j])) {
            val *= 10;
            val += arr[j] - '0';
            j ++;
        }
        //
        if(!operator.isEmpty() && isHighOp(operator.peek())) {
            double temp = getValue(res.pop(), val,operator.pop());
            val = temp;
        }
        if(j < arr.length && isHighOp(arr[j])) {
            res.push(val);
            operator.push(arr[j]);
        }
        else {
            double temp = getValue(res.pop(), val, operator.pop());
            res.push(temp);
            if(j < arr.length) operator.push(arr[j]);
        }
        i = j;
    }
    // notice the format;
    DecimalFormal df = new DecimalFormat("#.##");
    return Double.parseDouble(df.format(res.pop()));
}

public boolean isHighOp(Character ch) {
    return ch == '*' || ch == '/';
}

public double getValue(double op1, double op2, Character ch) {
    if(ch == '+') return op1 + op2;
    if(ch == '-') return op1 - op2;
    if(ch == '*') return op1 * op2;
    return op1 / op2;
}