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// 735. Asteroid Collision
#include "leetcode.h"
/*
* given an array 'asteroids' o intergers 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.
*/
class Solution {
public:
vector<int> asteroidCollision(vector<int> &asteroids) {
int n = asteroids.size();
stack<int> s;
for (int i = 0; i < n; i++) {
if (asteroids[i] || s.empty())
s.push(asteroids[i]);
else {
while (!s.empty() && s.top() && s.top() < abs(asteroids[i]))
s.pop();
if (!s.empty() && s.top() == abs(asteroids[i]))
s.pop();
else if (s.empty() || s.top() < 0)
s.push(asteroids[i]);
}
}
vector<int> ans(s.size());
for (int i = (int)s.size() - 1; i >= 0; i--) {
ans[i] = s.top();
s.pop();
}
return ans;
}
};
int main() {
Solution obj;
vector<int> a1 = {5, 10, -5}, a2 = {8, -8}, a3 = {10, 2, -5};
for (auto i : obj.asteroidCollision(a1))
printf("%d ", i); // expect: 5 10
printf("\n");
for (auto i : obj.asteroidCollision(a2))
printf("%d ", i); // expect: nil
printf("\n");
for (auto i : obj.asteroidCollision(a3))
printf("%d ", i); // expect: 10
printf("\n");
}
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