原题链接在这里：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
题解：
思路：1. 遇到数字位，看后一位是否为数字，若是位数字，当前位需要进十.
2. 开始设sign = 1，若遇到 ' - ', sign 改为 -1，若遇到 '+'，sign改回1.
3. 遇到 '(', 压栈，先压之前的res，后压sign，然后初始化res和sign.
4. 遇到')' ,出栈，当前res先乘pop() 出来的 sign，再加上pop()出来的之前结果.
5. Encountering space, just continue.
Note:1.遇到数字时需要生成一个新的cur 用来存储当前数，随后再用cur求res，不能直接求res, '-' 时会出问题.
Time Complexity: O(s.length()).
Space: O(s.length()), stack space.
AC Java:
1 public class Solution { 2 public int calculate(String s) { 3 if(s == null || s.length() == 0){ 4 return -1; 5 } 6 int res = 0; 7 int sign = 1; 8 Stack<Integer> stk = new Stack<Integer>(); 9 for(int i = 0; i<s.length(); i++){ 10 char c = s.charAt(i); 11 if(Character.isDigit(c)){ 12 int cur = (int)(c-'0'); 13 while(i+1 < s.length() && Character.isDigit(s.charAt(i+1))){ 14 cur = cur*10 + (int)(s.charAt(i+1) - '0'); 15 i = i+1; 16 } 17 res += sign*cur; 18 }else if(c == '-'){ 19 sign = -1; 20 }else if(c == '+'){ 21 sign = 1; 22 }else if(c == '('){ 23 stk.push(res); 24 res = 0; 25 stk.push(sign); 26 sign = 1; 27 }else if(c == ')'){ 28 res = res*stk.pop() + stk.pop(); 29 } 30 } 31 return res; 32 } 33 }