【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();
    }