Leetcode 735 planetary collision [stack simulation] the leetcode road of heroding

Their thinking ：
A more complex simulation problem , First of all, understand , The condition of planetary collision is that the speed of the planet on the left is positive , Only when the speed of the planets on the right is negative can they collide , And continuous collisions can occur , That is, the planet on the right is left and larger than the planet on the left and right , Collide until you encounter a planet larger than the one on the right, or the same size, or no planet on the right , So use stack to store . When traversing , First judge the direction , If you walk in the opposite direction ,while The cycle begins to collide , End of collision , Finally, if the opposite size is the same, they all disappear , There is no right planet to put the current planet into , Finally, traverse the stack and put vector And reverse , The code is as follows ：

class Solution {
    
public:
    vector<int> asteroidCollision(vector<int>& asteroids) {
    
        stack<int> s;
        vector<int> ans;
        for(auto& asteroid : asteroids) {
    
            if(s.empty()) {
    
                s.push(asteroid);
            } else {
    
                int top = s.top();
                if(top > 0 && asteroid < 0) {
    
                    while(top > 0 && asteroid + top < 0) {
    
                        s.pop();
                        if(s.empty()) {
    
                            break;
                        }
                        top = s.top();
                    }
                    if(top + asteroid == 0) {
    
                        s.pop();
                        continue;
                    }
                    if(s.empty() || (!s.empty() && s.top() < 0)) {
    
                        s.push(asteroid);
                    } 
                } else {
    
                    s.push(asteroid);
                }
            }
        }
        while(!s.empty()) {
    
            ans.push_back(s.top());
            s.pop();
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};