Formatted question description: https://leetcode.ca/all/735.html

735. Asteroid Collision (Medium)

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:

• 1 <= asteroids <= 104
• -1000 <= asteroids[i] <= 1000
• asteroids[i] != 0

Related Topics:
Stack

Similar Questions:

Solution 1.

• class Solution {
public int[] asteroidCollision(int[] asteroids) {
Stack<Integer> stack = new Stack<Integer>();
int length = asteroids.length;
for (int i = length - 1; i >= 0; i--) {
int asteroid = asteroids[i];
if (asteroid < 0)
stack.push(asteroid);
else {
boolean flag = true;
while (!stack.isEmpty() && stack.peek() < 0) {
int prev = stack.peek();
if (asteroid < -prev) {
flag = false;
break;
} else {
stack.pop();
if (asteroid == -prev) {
flag = false;
break;
}
}
}
if (flag)
stack.push(asteroid);
}
}
int size = stack.size();
int[] remain = new int[size];
for (int i = 0; i < size; i++)
remain[i] = stack.pop();
return remain;
}
}

############

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();
}
}


• // OJ: https://leetcode.com/problems/asteroid-collision/
// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> asteroidCollision(vector<int>& A) {
vector<int> ans;
for (int n : A) {
if (ans.empty() || ans.back() < 0 || n > 0) ans.push_back(n);
else {
while (ans.size() && ans.back() > 0 && ans.back() < -n) ans.pop_back();
if (ans.size() && ans.back() == -n) ans.pop_back();
else if (ans.empty() || ans.back() < 0) ans.push_back(n);
}
}
return ans;
}
};
};

• class Solution:
def asteroidCollision(self, asteroids: List[int]) -> List[int]:
ans = []
for a in asteroids:
if a > 0:
ans.append(a)
else:
while ans and 0 < ans[-1] < -a:
ans.pop()
if ans and ans[-1] == -a:
ans.pop()
elif not ans or ans[-1] < -a:
ans.append(a)
return ans

############

class Solution(object):
def asteroidCollision(self, asteroids):
"""
:type asteroids: List[int]
:rtype: List[int]
"""
stack = []
for ast in asteroids:
while stack and ast < 0 and stack[-1] >= 0:
pre = stack.pop()
if ast == -pre:
ast = None
break
elif -ast < pre:
ast = pre
if ast != None:
stack.append(ast)
return stack

• 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
}