【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
【解析】
直接拿测试用例走一下下面的程序,就会明吧其巧妙之处。
public class Solution {
public int calculate(String s) {
Stack<Integer> stack = new Stack<>();
stack.push(1);
stack.push(1);
int res = 0;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (Character.isDigit(c)) {
int num = c - '0';
int j = i + 1;
while (j < s.length() && Character.isDigit(s.charAt(j))) {
num = 10 * num + (s.charAt(j) - '0');
j++;
}
res += stack.pop() * num;
i = j - 1;
} else if (c == '+' || c == '(') {
stack.push(stack.peek());
} else if (c == '-') {
stack.push(-1 * stack.peek());
} else if (c == ')') {
stack.pop();
}
}
return res;
}
}