224. 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

链接： http://leetcode.com/problems/basic-calculator/

题解：

刚开始以为要用和RPN一样的方法，其实不是的。可以one pass遍历整个数组并且得到结果。 需要使用一个栈来cache括号这种情况。

Time Complexity - O(n)， Space Complexity - O(n)。

public class Solution {
public int calculate(String s) {
if(s == null || s.length() == 0)
return 0;
Stack<Integer> stack = new Stack<>();
int result = 0, curNum = 0, sign = 1;

for(int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if(c == ' ')
continue;
else if(Character.isDigit(c))
curNum = curNum * 10 + (int)(c - '0');
else if (c == '+') {
result += sign * curNum;
curNum = 0;
sign = 1;
} else if (c == '-') {
result += sign * curNum;
curNum = 0;
sign = -1;
} else if (c == '(') {
stack.push(result);
stack.push(sign);
curNum = 0;
sign = 1;
result = 0;
} else if (c == ')') {
result += sign * curNum;
curNum = 0;
if(!stack.isEmpty())
result *= stack.pop();
if(!stack.isEmpty())
result += stack.pop();
}
}

if(curNum != 0)
result += sign * curNum;

return result;
}
}

Reference:
https://leetcode.com/discuss/39553/iterative-java-solution-with-stack https://leetcode.com/discuss/41868/java-solution-stack https://leetcode.com/discuss/39479/simple-c-in-24-ms