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This approach uses a stack to store numbers and consider operator precedence. Multiplication and division are processed immediately, whereas addition and subtraction modify how numbers are pushed onto the stack.
Time Complexity: O(n) where n is the length of the string. Space Complexity: O(n) for storing numbers in the stack.
1#include <stdio.h>
2#include <stdlib.h>
3#include <ctype.h>
4
5int calculate(char *s) {
6 int length = strlen(s);
7 int stack[length];
8 int top = -1;
9 char sign = '+';
10 int num = 0;
11 for (int i = 0; i < length; i++) {
12 if (isdigit(s[i])) {
13 num = num * 10 + (s[i] - '0');
14 }
15 if ((!isdigit(s[i]) && !isspace(s[i])) || i == length - 1) {
16 if (sign == '+') {
17 stack[++top] = num;
18 } else if (sign == '-') {
19 stack[++top] = -num;
20 } else if (sign == '*') {
21 stack[top] *= num;
22 } else if (sign == '/') {
23 stack[top] /= num;
24 }
25 sign = s[i];
26 num = 0;
27 }
28 }
29
30 int result = 0;
31 while (top != -1) {
32 result += stack[top--];
33 }
34 return result;
35}
36
37int main() {
38 char expression[] = "3+2*2";
39 printf("Output: %d\n", calculate(expression));
40 return 0;
41}This C solution iterates over the string, calculates numeric values, updates the stack with results immediately for '*', '/', while '+' and '-' add or subtract the number to/from the stack respectively. Finally, calculates the sum of values in the stack.
This approach parses the expression twice, first to handle multiplications and divisions, then to process additions and subtractions. This avoids using a stack and simplifies sign handling.
Time Complexity: O(n) where n is the number of characters in the input string. Space Complexity: O(1) since it only uses primitive data types.
1public
This Java solution focuses on efficiency by keeping track of only necessary information for computation: current number, last computed number, and current result. It simplifies calculations avoiding extra space usage.