README_EN.md

735. Asteroid Collision

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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.

Example 4:

Input: asteroids = [-2,-1,1,2]
Output: [-2,-1,1,2]
Explanation: The -2 and -1 are moving left, while the 1 and 2 are moving right. Asteroids moving the same direction never meet, so no asteroids will meet each other.

 

Constraints:

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

Solutions

this can be analogous to matching parentheses:

• right-moving asteroid (left parenthesis): will not cause a collision, will be pushed directly
• left-moving asteroid (right parenthesis): may collision with the right-moving asteroid, special treatment

because the answer requires all the asteroids left after the collision, which is element left in the stack, you can simply represent the stack in an array

Python3

class Solution:
    def asteroidCollision(self, asteroids: List[int]) -> List[int]:
        ans = []
        for a in asteroids:
            if a > 0:
                ans.append(a)
            else:
                while len(ans) > 0 and ans[-1] > 0 and ans[-1] < -a:
                    ans.pop()
                if len(ans) > 0 and ans[-1] == -a:
                    ans.pop()
                elif len(ans) == 0 or ans[-1] < -a:
                    ans.append(a)
        return ans

Java

class Solution {
    public int[] asteroidCollision(int[] asteroids) {
        Deque<Integer> d = new ArrayDeque<>();
        for (int a : asteroids) {
            if (a > 0) {
                d.offerLast(a);
            } else {
                while (!d.isEmpty() && d.peekLast() > 0 && d.peekLast() < -a) {
                    d.pollLast();
                }
                if (!d.isEmpty() && d.peekLast() == -a) {
                    d.pollLast();
                } else if (d.isEmpty() || d.peekLast() < -a) {
                    d.offerLast(a);
                }
            }
        }
        return d.stream().mapToInt(Integer::valueOf).toArray();
    }
}

C++

class Solution {
public:
    vector<int> asteroidCollision(vector<int>& asteroids) {
        vector<int> ans;
        for (int a : asteroids) {
            if (a > 0) {
                ans.push_back(a);
            } else {
                while (!ans.empty() && ans.back() > 0 && ans.back() < -a) {
                    ans.pop_back();
                }
                if (!ans.empty() && ans.back() == -a) {
                    ans.pop_back();
                } else if (ans.empty() || ans.back() < -a) {
                    ans.push_back(a);
                }
            }
        }
        return ans;
    }
};

Go

func asteroidCollision(asteroids []int) []int {
	var ans []int
	for _, a := range asteroids {
		if a > 0 {
			ans = append(ans, a)
		} else {
			for len(ans) > 0 && ans[len(ans)-1] > 0 && ans[len(ans)-1] < -a {
				ans = ans[:len(ans)-1]
			}
			if len(ans) > 0 && ans[len(ans)-1] == -a {
				ans = ans[:len(ans)-1]
			} else if len(ans) == 0 || ans[len(ans)-1] < -a {
				ans = append(ans, a)
			}
		}
	}
	return ans
}

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