Basic Calculator
Total Accepted: 26713 Total Submissions: 121377 Difficulty: Hard
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
- 不难,但是还挺容易错误的
- 参考图文解释
- 最直观的想法就是一个stack存数字,一个stack 存 () +-这些符号
- 解法用存入同一个stack 1, -1 来模拟了加减
public class Solution {
public int calculate(String s) {
Stack<Integer> stackInt = new Stack<>();
stackSign = new Stack<>();
int result = 0;
int sign = 1;
int num = 0;
for(int i=0; i<s.length(); i++) {
char c = s.charAt(i);
if(c >= '0' && c <= '9') {
int digit = Character.getNumericValue(c);
num = 10 * num + digit;
} else if(c == '+' || c == '-') {
result += num * sign;
num = 0;
sign = c == '+' ? 1 : -1;
} else if(c == '(') {
stackInt.push(result);
stackSign.push(sign);
result = 0;
sign = 1;
} else if(c == ')') {
result += num * sign;
result = result * stackSign.pop() + stackInt.pop();
num = 0;
sign = 1;
}
}
result += num * sign;
return result;
}
}