| comments | difficulty | edit_url |
|---|---|---|
true |
Medium |
Given an arithmetic equation consisting of positive integers, +, -, * and / (no parentheses), compute the result.
The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.
Example 1:
Input: "3+2*2"
Output: 7
Example 2:
Input: " 3/2 "
Output: 1
Example 3:
Input: " 3+5 / 2 "
Output: 5
Note:
- You may assume that the given expression is always valid.
- Do not use the eval built-in library function.
We can use a stack to store numbers. Each time we encounter an operator, we push the number into the stack. For addition and subtraction, since their priority is the lowest, we can directly push the numbers into the stack. For multiplication and division, since their priority is higher, we need to take out the top element of the stack, perform multiplication or division with the current number, and then push the result back into the stack.
Finally, the sum of all elements in the stack is the answer.
The time complexity is
class Solution:
def calculate(self, s: str) -> int:
n = len(s)
x = 0
sign = "+"
stk = []
for i, c in enumerate(s):
if c.isdigit():
x = x * 10 + ord(c) - ord("0")
if i == n - 1 or c in "+-*/":
match sign:
case "+":
stk.append(x)
case "-":
stk.append(-x)
case "*":
stk.append(stk.pop() * x)
case "/":
stk.append(int(stk.pop() / x))
x = 0
sign = c
return sum(stk)class Solution {
public int calculate(String s) {
int n = s.length();
int x = 0;
char sign = '+';
Deque<Integer> stk = new ArrayDeque<>();
for (int i = 0; i < n; ++i) {
char c = s.charAt(i);
if (Character.isDigit(c)) {
x = x * 10 + (c - '0');
}
if (i == n - 1 || !Character.isDigit(c) && c != ' ') {
switch (sign) {
case '+' -> stk.push(x);
case '-' -> stk.push(-x);
case '*' -> stk.push(stk.pop() * x);
case '/' -> stk.push(stk.pop() / x);
}
x = 0;
sign = c;
}
}
int ans = 0;
while (!stk.isEmpty()) {
ans += stk.pop();
}
return ans;
}
}class Solution {
public:
int calculate(string s) {
int n = s.size();
int x = 0;
char sign = '+';
stack<int> stk;
for (int i = 0; i < n; ++i) {
char c = s[i];
if (isdigit(c)) {
x = x * 10 + (c - '0');
}
if (i == n - 1 || !isdigit(c) && c != ' ') {
if (sign == '+') {
stk.push(x);
} else if (sign == '-') {
stk.push(-x);
} else if (sign == '*') {
int y = stk.top();
stk.pop();
stk.push(y * x);
} else if (sign == '/') {
int y = stk.top();
stk.pop();
stk.push(y / x);
}
x = 0;
sign = c;
}
}
int ans = 0;
while (!stk.empty()) {
ans += stk.top();
stk.pop();
}
return ans;
}
};func calculate(s string) (ans int) {
n := len(s)
x := 0
sign := '+'
stk := []int{}
for i := range s {
if s[i] >= '0' && s[i] <= '9' {
x = x*10 + int(s[i]-'0')
}
if i == n-1 || (s[i] != ' ' && (s[i] < '0' || s[i] > '9')) {
switch sign {
case '+':
stk = append(stk, x)
case '-':
stk = append(stk, -x)
case '*':
stk[len(stk)-1] *= x
case '/':
stk[len(stk)-1] /= x
}
x = 0
sign = rune(s[i])
}
}
for _, x := range stk {
ans += x
}
return
}function calculate(s: string): number {
const n = s.length;
let x = 0;
let sign = '+';
const stk: number[] = [];
for (let i = 0; i < n; ++i) {
if (!isNaN(Number(s[i])) && s[i] !== ' ') {
x = x * 10 + s[i].charCodeAt(0) - '0'.charCodeAt(0);
}
if (i === n - 1 || (isNaN(Number(s[i])) && s[i] !== ' ')) {
switch (sign) {
case '+':
stk.push(x);
break;
case '-':
stk.push(-x);
break;
case '*':
stk.push(stk.pop()! * x);
break;
default:
stk.push((stk.pop()! / x) | 0);
}
x = 0;
sign = s[i];
}
}
return stk.reduce((x, y) => x + y);
}class Solution {
func calculate(_ s: String) -> Int {
let n = s.count
var x = 0
var sign: Character = "+"
var stk = [Int]()
let sArray = Array(s)
for i in 0..<n {
let c = sArray[i]
if c.isNumber {
x = x * 10 + Int(String(c))!
}
if i == n - 1 || (!c.isNumber && c != " ") {
switch sign {
case "+":
stk.append(x)
case "-":
stk.append(-x)
case "*":
if let last = stk.popLast() {
stk.append(last * x)
}
case "/":
if let last = stk.popLast() {
stk.append(last / x)
}
default:
break
}
x = 0
sign = c
}
}
return stk.reduce(0, +)
}
}