LeetCode-in-Java
224. Basic Calculator
Hard
Given a string s representing a valid expression, implement a basic calculator to evaluate it, and return the result of the evaluation.
Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().
Example 1:
Input: s = “1 + 1”
Output: 2
Example 2:
Input: s = “ 2-1 + 2 “
Output: 3
Example 3:
Input: s = “(1+(4+5+2)-3)+(6+8)”
Output: 23
Constraints:
1 <= s.length <= 3 * 105sconsists of digits,'+','-','(',')', and' '.srepresents a valid expression.'+'is not used as a unary operation (i.e.,"+1"and"+(2 + 3)"is invalid).'-'could be used as a unary operation (i.e.,"-1"and"-(2 + 3)"is valid).- There will be no two consecutive operators in the input.
- Every number and running calculation will fit in a signed 32-bit integer.
Solution
public class Solution {
private int i = 0;
public int calculate(String s) {
char[] ca = s.toCharArray();
return helper(ca);
}
private int helper(char[] ca) {
int num = 0;
int prenum = 0;
boolean isPlus = true;
for (; i < ca.length; i++) {
char c = ca[i];
if (c != ' ') {
if (c >= '0' && c <= '9') {
if (num == 0) {
num = (c - '0');
} else {
num = num * 10 + c - '0';
}
} else if (c == '+') {
prenum += num * (isPlus ? 1 : -1);
isPlus = true;
num = 0;
} else if (c == '-') {
prenum += num * (isPlus ? 1 : -1);
num = 0;
isPlus = false;
} else if (c == '(') {
i++;
prenum += helper(ca) * (isPlus ? 1 : -1);
isPlus = true;
num = 0;
} else if (c == ')') {
return prenum + num * (isPlus ? 1 : -1);
}
}
}
return prenum + num * (isPlus ? 1 : -1);
}
}