Force deduction question 3

735. Planetary collision

Medium difficulty 305 Switch to English to receive dynamic feedback

Given an array of integers  asteroids, Represents planets on the same line .

For each element in the array , Its absolute value represents the size of the planet , Positive and negative indicate the direction of movement of the planet （ Positive means moving to the right , Negative means moving to the left ）. Every planet moves at the same speed .

Find all the planets left after the collision . Collision rules ： The two planets collide with each other , Smaller planets will explode . If two planets are the same size , Then both planets will explode . Two planets moving in the same direction , There will never be a collision .

Example 1：

 Input ：asteroids = [5,10,-5]
 Output ：[5,10]
 explain ：10  and  -5  After the collision, only  10 . 5  and  10  There will never be a collision .

Example 2：

 Input ：asteroids = [8,-8]
 Output ：[]
 explain ：8  and  -8  After the collision , Both exploded .

Example 3：

 Input ：asteroids = [10,2,-5]
 Output ：[10]
 explain ：2  and  -5  After the collision, there are  -5 .10  and  -5  After the collision, there are  10 .

Tips ：

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

class Solution {
    public int[] asteroidCollision(int[] asteroids) {
        // Utilization stack , Positive numbers are stored inside , Encountered a negative number stack 
        Stack<Integer> st = new Stack<>();
        for(int i = 0; i < asteroids.length; i++){
            if(asteroids[i] > 0) st.push(asteroids[i]);
            else if((asteroids[i] < 0 && st.isEmpty()) || st.peek() < 0) st.push(asteroids[i]);
            else {
                int pop = st.pop();
                if((pop + asteroids[i]) > 0)st.push(pop);
                else if((pop + asteroids[i]) < 0){
                    i--;
                }
            }
        }
        Object[] ob = st.toArray();
        int[] nums = new int[st.size()];
        for(int i = 0; i < ob.length; i++){
            nums[i] = (int)ob[i];
        }
        return nums;
    }
}