Description
Given a string s which represents an expression, evaluate this expression and return its value.
The integer division should truncate toward zero.
You may assume that the given expression is always valid. All intermediate results will be in the range of [-231, 231 - 1].
💡 Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().
Example 1:
Input: s = "3+2*2"
Output: 7
Example 2:
Input: s = " 3/2 "
Output: 1
Example 3:
Input: s = " 3+5 / 2 "
Output: 5
Constraints:
1 <= s.length <= 3 * 105-
sconsists of integers and operators('+', '-', '*', '/')separated by some number of spaces. -
srepresents a valid expression. - All the integers in the expression are non-negative integers in the range
[0, 231 - 1]. - The answer is guaranteed to fit in a 32-bit integer.
Solutions
Solution 1
Intuition
use a variety to store last operation, if found last operation is * or / pop it and calculate it with current value;
Code
import java.util.Stack;
/*
* @lc app=leetcode id=227 lang=java
*
* [227] Basic Calculator II
*/
// @lc code=start
class Solution {
public int calculate(String s) {
Stack<Integer> stack = new Stack<>();
int current = 0;
char lastOperation = '+';
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (Character.isDigit(c)) {
current = current * 10 + c - '0';
}
if (!Character.isDigit(c) && !Character.isWhitespace(c) || i == s.length() - 1) {
switch (lastOperation) {
case '+':
stack.push(current);
break;
case '-':
stack.push(-current);
break;
case '*':
stack.push(stack.pop() * current);
break;
case '/':
stack.push(stack.pop() / current);
break;
}
lastOperation = c;
current = 0;
}
}
int sum = 0;
for (Integer integer : stack) {
sum += integer;
}
return sum;
}
}
// @lc code=end
Complexity
- Time: O(n)
- Space: O(n)
Solution 2
Intuition
Code
Complexity
- Time:
- Space:
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