LeetCode Basic Calculator

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:

 public class Solution {
     public int calculate(String s) {
         if(s == null || s.length() == 0){
             return -1;
         }
         int res = 0;
         int sign = 1;
         Stack<Integer> stk = new Stack<Integer>();
         for(int i = 0; i<s.length(); i++){
             char c = s.charAt(i);
             if(Character.isDigit(c)){
                 int cur = (int)(c-'0');
                 while(i+1 < s.length() && Character.isDigit(s.charAt(i+1))){
                     cur = cur*10 + (int)(s.charAt(i+1) - '0');
                     i = i+1;
                 }
                 res += sign*cur;
             }else if(c == '-'){
                 sign = -1;
             }else if(c == '+'){
                 sign = 1;
             }else if(c == '('){
                 stk.push(res);
                 res = 0;
                 stk.push(sign);
                 sign = 1;
             }else if(c == ')'){
                 res = res*stk.pop() + stk.pop();
             }
         }
         return res;
     }
 }

类似Evaluate Reverse Polish Notation.

跟上Basic Calculator II.