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
public int calculate(String s) {
if (s==null || s.length() == 0) return 0;
Stack<Integer> st = new Stack<Integer>();
int res = 0, sign = 1, len = s.length();
for (int i = 0; i < len; i++) {
char c = s.charAt(i);
if (Character.isDigit(c)) {
int cur = c - '0';
while (i + 1 < len && Character.isDigit(s.charAt(i + 1))) {
cur = 10 * cur + s.charAt(++i) - '0';
}
res += sign * cur;
}
else if (c == '-') sign = -1;
else if (c == '+') sign = 1;
else if (c == '(') {
st.push(res);
res = 0;
st.push(sign);
sign = 1;
}
else if (c == ')') {
res = st.pop() * res + st.pop();
sign = 1;
}
}
return res;
Implement a basic calculator to evaluate a simple expression string.
The expression string contains only non-negative integers, +, -, *, / operators
and empty spaces . The integer division should truncate toward zero.
You may assume that the given expression is always valid.
Some examples:
"3+2*2" = 7 " 3/2 " = 1 " 3+5 / 2 " = 5
public int calculate(String s) {
Stack<Integer> stack = new Stack<>();
char sign = '+';
int len = s.length();
int n = 0;
for (int i = 0; i < len; i++) {
char c = s.charAt(i);
if (Character.isDigit(c)) {
n = n * 10 + c - '0';
}
if (!Character.isDigit(c) && c != ' ' || i == len - 1) {
if (sign == '+') {
stack.push(n);
} else if (sign == '-') {
stack.push(-n);
} else if (sign == '*') {
stack.push(stack.pop() * n);
} else if (sign == '/') {
stack.push(stack.pop() / n);
}
sign = c;
n = 0;
}
}
int re = 0;
while (!stack.isEmpty()) re += stack.pop();
return re;
}
本文介绍了一种简单表达式计算器的实现方法,该计算器能够处理包含括号、加减乘除运算符及空格的非负整数表达式,并正确进行运算。文中提供了两个示例代码:一个用于处理加减法及括号的表达式;另一个则能处理包括乘除在内的四则运算。
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