Day trip from Lianjiang to Haikou
Haikou lies 179 km (111 mi) south of Lianjiang, in the same time zone. What the numbers say about getting there.
At 179 km (111 mi) apart, Haikou is a realistic day trip from Lianjiang: about 2 h 55 min each way by road.
| Lianjiang | Haikou | |
|---|---|---|
| Country | China | China |
| Population | 1.4 million | 3 million |
| Elevation | 46 m (151 ft) | 9 m (30 ft) |
| Time now | UTC+8 (Asia/Shanghai) | UTC+8 (Asia/Shanghai) |
| Postal codes | 524400 | 570000 |
Where to book tours and tickets in Haikou
We take no bookings for these providers and hold no reservation with them. The provider you choose handles payment, confirmation, changes and refunds. How that works.
Staying in Haikou
Our booking partner lists 113 places to stay in Haikou. Of the 107 with a star rating, the largest group is 3-star (39), and 20 are five-star. Guests have left 1,410 reviews in total.
Book a hotel in Haikou
Live rooms and prices from our booking partner, booked and paid for here. Free cancellation is marked where it applies.
Common questions
How far is Haikou from Lianjiang?
179 km (111 mi) in a straight line, south of Lianjiang; by road roughly 219 km (136 mi).
How long does it take to get from Lianjiang to Haikou by road?
About 2 h 55 min, estimated from distance and typical speeds. Timetables and traffic decide the real figure.
Is there a time difference between Lianjiang and Haikou?
No, both are on Asia/Shanghai at the moment.
Where can I stay in Haikou?
Our booking partner lists 113 places to stay in Haikou, bookable on this site.
Where these facts come from
- GeoNames (CC BY 4.0): population, elevation, time zone, postal codes
- Wikidata (CC0): area, sister cities, visitor numbers, heritage listings, stations and ports
- OurAirports (public domain): airports, runways
- Natural Earth (public domain): map outlines
Data compiled 14 September 2026. Hotel and hostel figures are from our booking partner’s catalogue on that date. Journey times on this page are estimates from distance, not timetables; check the operator before you rely on one.