LeetCode //C - 227. Basic Calculator II

227. Basic Calculator II

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 $[-2^{31},2^{31}-1]$.

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 * $10^5$
• s consists of integers and operators (‘+’, ‘-’, ‘*’, ‘/’) separated by some number of spaces.
• s represents a valid expression.
• All the integers in the expression are non-negative integers in the range $[0,2^{31}-1]$.
• The answer is guaranteed to fit in a 32-bit integer.

From: LeetCode
Link: 227. Basic Calculator II

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Solution:

Ideas:

• Stack Initialization: int stack[300000]; - We use a large array to simulate the stack.
• Variables: int top = -1; - This variable keeps track of the top of the stack. int currentNumber = 0; - This stores the current number being processed.
• Iteration through string: The loop iterates through each character in the string.
  • If a digit is found, it builds the currentNumber.
  • If a non-digit or end of string is found, the code processes the previous number with the current operation.
• Operations Handling:
  • Based on the operation, the code either pushes the number onto the stack or performs multiplication/division and updates the stack.
• Final Calculation: The loop sums all the values in the stack to get the final result.

Code:

int calculate(char* s) {
    int stack[300000];
    int top = -1;
    int currentNumber = 0;
    char operation = '+';
    
    for (int i = 0; s[i] != '\0'; i++) {
        if (isdigit(s[i])) {
            currentNumber = currentNumber * 10 + (s[i] - '0');
        }
        if (!isdigit(s[i]) && !isspace(s[i]) || s[i + 1] == '\0') {
            if (operation == '+') {
                stack[++top] = currentNumber;
            } else if (operation == '-') {
                stack[++top] = -currentNumber;
            } else if (operation == '*') {
                stack[top] *= currentNumber;
            } else if (operation == '/') {
                stack[top] /= currentNumber;
            }
            operation = s[i];
            currentNumber = 0;
        }
    }
    
    int result = 0;
    for (int i = 0; i <= top; i++) {
        result += stack[i];
    }
    
    return result;
}

原文地址：https://blog.csdn.net/navicheung/article/details/140428343

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