LeetCode 2211. Count Collisions on a Road
There are?n?cars on an infinitely long road. The cars are numbered from?0?to?n - 1?from left to right and each car is present at a?unique?point.
You are given a?0-indexed?string?directions?of length?n.?directions[i]?can be either?'L',?'R', or?'S'?denoting whether the?ith?car is moving towards the?left, towards the?right, or?staying?at its current point respectively. Each moving car has the?same speed.
The number of collisions can be calculated as follows:
When two cars moving in?opposite?directions collide with each other, the number of collisions increases by?
2.When a moving car collides with a stationary car, the number of collisions increases by?
1.
After a collision, the cars involved can no longer move and will stay at the point where they collided. Other than that, cars cannot change their state or direction of motion.
Return?the?total number of collisions?that will happen on the road.
?
Example 1:
Input: directions = "RLRSLL"
Output: 5
Explanation:The collisions that will happen on the road are:?
- Cars 0 and 1 will collide with each other. Since they are moving in opposite directions, the number of collisions becomes 0 + 2 = 2.
- Cars 2 and 3 will collide with each other. Since car 3 is stationary, the number of collisions becomes 2 + 1 = 3.?
- Cars 3 and 4 will collide with each other. Since car 3 is stationary, the number of collisions becomes 3 + 1 = 4.?
- Cars 4 and 5 will collide with each other.?
After car 4 collides with car 3, it will stay at the point of collision and get hit by car 5. The number of collisions becomes 4 + 1 = 5.
Thus, the total number of collisions that will happen on the road is 5.
Example 2:
Input: directions = "LLRR"
Output: 0
Explanation:No cars will collide with each other. Thus, the total number of collisions that will happen on the road is 0.
?
Constraints:
1 <= directions.length <= 105directions[i]?is either?'L',?'R', or?'S'.這個是過了代碼，先說說不過的代碼，將RL的變成SS，然后就是去判斷每個位置最近的S在哪里，L的判斷左邊的S，R的判斷右邊的S，然后就Time Limit Exceed了。
仔細想想，直接先將RL的變成SS，然后找到最左端的S的位置，跟最右端S的位置，只要是L的位置小于最左端S的位置，R的位置小于最右端S的位置，那么一定會碰撞的。然后就過了。
Runtime:?41 ms, faster than?50.00%?of?Java?online submissions for?Count Collisions on a Road.
Memory Usage:?44.2 MB, less than?15.85%?of?Java?online submissions for?Count Collisions on a Road.
