735. Asteroid-Collision

difficulty: Medium

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]..

Method One

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

这个题的 edge cases 非常多，写一个没毛病的代码非常不容易。 一个观察是 如果空栈 或者 正数 或者 堆顶是负数 直接无脑入栈。 这样保证我们处理进入的是一个 负数。 任何时候只要发现 空栈 或者 堆顶元素是负数，直接终止。 我们用一个 flag 标记进入的元素是否被摧毁。

class Solution {
    public int[] asteroidCollision(int[] asteroids) {
        ArrayDeque<Integer> stack = new ArrayDeque<>();
        for (int i = 0; i < asteroids.length; i++) {
            if (stack.isEmpty() || asteroids[i] > 0 || stack.peek() < 0) {
                stack.push(asteroids[i]);
                continue;
            }
            boolean isDestroyed = false;
            while (!stack.isEmpty() && stack.peek() > 0) {
                int top = stack.peek();
                if (top < Math.abs(asteroids[i])) {
                    stack.pop();
                } else if (top == Math.abs(asteroids[i])) {
                    isDestroyed = true;
                    stack.pop();
                    break;
                } else {
                    isDestroyed = true;
                    break;
                }
            }
            if (!isDestroyed) {
                stack.push(asteroids[i]);
            }
        }
        int[] ans = new int[stack.size()];
        for (int i = ans.length - 1; i >= 0 ; i--) {
            ans[i] = stack.pop();
        }
        return ans;
    }
}