Hex.*

LC735 - Asteroid Collision

Posted on In
Leetcode: 155/2658 (59+76+20)

Description

We are given an array asteroids of integers representing asteroids in a row.
For each asteroid, the absolute value represents its size, and the sign represents its direction (positive meaning right, negative meaning left). Each asteroid moves at the same speed.
Find out the state of the asteroids after all collisions. If two asteroids meet, the smaller one will explode. If both are the same size, both will explode. Two asteroids moving in the same direction will never meet.

Example 1:
Input: asteroids = [5,10,-5]
Output: [5,10]
Explanation: The 10 and -5 collide resulting in 10. The 5 and 10 never collide.

Example 2:
Input: asteroids = [8,-8]
Output: []
Explanation: The 8 and -8 collide exploding each other.

Example 3:
Input: asteroids = [10,2,-5]
Output: [10]
Explanation: The 2 and -5 collide resulting in -5. The 10 and -5 collide resulting in 10.

Constraints:
2 <= asteroids.length <= 104
-1000 <= asteroids[i] <= 1000
asteroids[i] != 0

Solution

Since collision cancellation always occurs between adjacent planets, we can use “stacks” to simulate this process.

Process all the ats[i] from the front to the back, and use the stack to store the planets that are not currently offset. When the direction of the top element of the stack is to the right, and the current ats[i]ats[i] direction is to the left, the offset will occur. Operation, the offset process can be carried out according to the rules.

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# O(n) time | O(n) space
class Solution:
    def asteroidCollision(self, asteroids: List[int]) -> List[int]:
        ans = []
        for star in asteroids:
            while ans and ans[-1] > 0 and star < 0:
                pre, cur = ans[-1], -star
                if pre <= cur:
                    ans.pop()
                if pre >= cur:
                    break
            else:
                ans.append(star)
        return ans