735. Asteroid Collision

Description

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].
Problem URL

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Solution

给一个数组，绝对值表示小行星大小，正负代表方向，小的撞到大的gg，同一方向的不会碰到。问稳定之后的小行星情况。

Using a stack to store result. If stack is empty or a[I] > 0, simply push it into stack. Else, a[I] is < 0, we use peek() element in stack to compare with a[I]. If prev < 0, push a[I] into stack, it has same direction; If prev == -a[I], pop it, break, collide; If prev > -a[I], a[I] collide, just break; or prev explode, continue pop;

Code

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

Time Complexity: O(n^2)
Space Complexity: O(n)

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