题解 | #逆波兰表达式求值#

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param tokens string字符串一维数组
 * @param tokensLen int tokens数组长度
 * @return int整型
 */
int evalRPN(char** tokens, int tokensLen ) {
    // write code here
    int stack[tokensLen];
    int top = 0;
    for (int i = 0; i < tokensLen; i++) {
        char* num = tokens[i];
        if ( strlen(num) > 1 || '0' <= num[0] && num[0] <= '9') {
            stack[top++] = atoi(num);
        } else {
            switch (*num) {
                case '+':
                    stack[top - 2] = stack[top - 2] + stack[top - 1];
                    top--;
                    break;
                case '-':
                    stack[top - 2] = stack[top - 2] - stack[top - 1];
                    top--;
                    break;
                case '*':
                    stack[top - 2] = stack[top - 2] * stack[top - 1];
                    top--;
                    break;
                case '/':
                    stack[top - 2] = stack[top - 2] / stack[top - 1];
                    top--;
                    break;
            }
        }
    }
    return stack[top - 1];
}