【Leetcode】Basic Calculator

题目链接：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
Note: Do not use the eval built-in library function.

思路：
用栈保存操作数、操作符，要注意处理右括号的情况。

算法：

    public int calculate(String s) {        Stack<String> operator = new Stack<String>();        Stack<Integer> nums = new Stack<Integer>();        s = s.replace(" ", "");        int curNum = 0; // 连续字符都是数字要组合成一个数        for (int i = 0; i < s.length(); i++) {            char op = s.charAt(i);            if (Character.isDigit(op)) { // 当前是数字                curNum = Integer.parseInt(op + "");                while (i + 1 < s.length() && Character.isDigit(s.charAt(i + 1))) { // 后继字符也是数字，组合成一个数                    char nextNum = s.charAt(i + 1);                    curNum = curNum * 10 + Integer.parseInt(nextNum + "");                    i++;                }// 一个操作数处理完                if (!operator.isEmpty()                        && nums.size() >= 1                        && (operator.peek().equals("+") || operator.peek()                                .equals("-"))) { // 若操作数栈有操作数，且                                                    // 操作符栈有+/-，则将当前数和栈中的数做相应处理                    String tmp = operator.pop();                    if (tmp.equals("-")) {                        int first = nums.pop();                        nums.push(first - curNum);                    } else {                        nums.push(nums.pop() + curNum);                    }                } else {                    nums.push(curNum);                }            } else { // 当前字符是操作符                if (op == '+' || op == '-' || op == '(') {                    operator.push(op + "");                } else if (op == ')') {                    if (operator.peek().equals("(")) { // 若括号消除，则要处理左括号前面的运算，即处理                                                        // 1+(4)中+的计算                        operator.pop();                        if (!operator.isEmpty()                                && nums.size() >= 2                                && (operator.peek().equals("+") || operator                                        .peek().equals("-"))) {                            String tmp = operator.pop();                            if (tmp.equals("-")) {                                int first = nums.pop();                                int second = nums.pop();                                nums.push(second - first);                            } else {                                nums.push(nums.pop() + nums.pop());                            }                        }                    } else {                        operator.push(")");                    }                }            }        }        return nums.pop();    }