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.

Note:

• The length of asteroids will be at most 10000.
• Each asteroid will be a non-zero integer in the range [-1000, 1000]..

Solution

分析题目后发现规律，每一次新扫描的数都需要与之前扫过的最后一个数进行比较，按照题目场景进行处理：

1. 如果是正数，肯定不会撞，直接记录入结果。
2. 如果是负数，有可能相撞，需要进行判断：
   • previous < current （注意是绝对值）： 则previous销毁。但需注意，销毁以后，它之前的数也可能被销毁，所以要一直向前比较，直到不能再销毁为止。Eg. [5, 2, 2, 2, -5]，-5 会把前面所有的2都销毁掉 => [5, -5]
   • 前一步判定完成后，还需要继续判断, previous == current , 则大家都销毁: => []
   • 或者，如果previous > current : 则current销毁。
   • 但如果此时前面所有都销毁了，已经是[]，或者前面是 负数，则不可能相撞，直接记录入结果。

class Solution {
    public int[] asteroidCollision(int[] asteroids) {
        if (asteroids == null || asteroids.length == 0) {
            return null;
        }
        
        Stack <Integer> tracker = new Stack <> ();
        
        for (int i = 0; i < asteroids.length; i++) {
            if (asteroids[i] > 0) {
                tracker.push (asteroids [i]);
            }
            else {
                while (!tracker.isEmpty () && tracker.peek () > 0 && tracker.peek () < -asteroids [i])
                {
                    tracker.pop ();
                }
                
                if (!tracker.isEmpty () && tracker.peek () == -asteroids [i]) {
                    tracker. pop ();
                    continue;
                }
                else if (tracker.isEmpty () || tracker.peek () < 0) { // all crashed or the previous one is "-"
                    tracker.push (asteroids [i]);
                }
            }
        }
        
        return getArray (tracker);
    }
    
    private int [] getArray (Stack <Integer> tracker)
    {
        int i = 0;
        int [] result = new int [tracker.size ()];
        
        for (int entry : tracker) {
            result [i ++] = entry;
        }
        
        return result;
    }
}