[LeetCode ] Basic Calculator

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Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces .

Example 1:

Input: "1 + 1"
Output: 2

Example 2:

Input: " 2-1 + 2 "
Output: 3

Example 3:

Input: "(1+(4+5+2)-3)+(6+8)"
Output: 23

Note:

• You may assume that the given expression is always valid.
• Do not use the eval built-in library function

题意：给出一个只有+-()和非负数的合法表达式，计算这个表达式的值。

思路：大体思路是每次遇到的如果不是），就入栈，遇到），执行出栈操作，直到遇到（，出栈过程中结算这个括号里的表达式的值，然后把这个值在入栈，写起来细节还是挺多的。

C代码：

int calculate(char* s) {
    int i,len,top,j;
    long long stack[100005],temp1,temp2,res;
    const long long op1 = (long long)INT_MAX + 1,op2 = op1 + 1,op3 = op2 + 1;
    res = top = 0;
    len = strlen(s);
    for(i = 0; i < len; i++) {
        if(s[i] == '(') {
            stack[top++] = op3;
        }
        else if(s[i] == '+') {
            stack[top++] = op2;
        }
        else if(s[i] == '-') {
            stack[top++] = op1;
        }
        else if(s[i] >= '0' && s[i] <= '9') {
            temp1 = 0;
            for(j = i; s[j] >= '0' && s[j] <= '9'; j++) {
                temp1 = temp1 * 10 + s[j] - '0';
            }
            stack[top++] = temp1;
            i = j - 1;
        }
        else if(s[i] == ')') {
            temp1 = 0;
            temp2 = 0;
            top--;
            while(stack[top] != op3) {
                if(stack[top] < op1) {
                    temp1 = stack[top];
                }
                else {
                    if(stack[top] == op1) temp2 -= temp1;
                    else if(stack[top] == op2) temp2 += temp1;
                }
                top--;
            }
            stack[top++] = temp2 + temp1;
        }
    }
    int flag = 1;
    for(i = 0; i < top; i++) {
        if(stack[i] == op2) flag = 1;
        else if(stack[i] == op1) flag = -1;
        else res += flag * stack[i];
    }
    return res;
}