735 Asteroid Collision

735. Asteroid Collision

1. Question

We are given an arrayasteroidsof 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 ofasteroidswill be at most10000.

Each asteroid will be a non-zero integer in the range[-1000, 1000]..

2. Implementation

(1) Stack

思路: 题目已经列明好几个规则，归结于三类:

• 如果两个行星方向一样，或者一个向左，一个向右，则不会发生碰撞

• 如果两个行星碰撞，size较小的那个会消失

• 如果两个size相同的行星碰撞，两个行星都会消失

因为会涉及到之前保留的行星到后面遇到碰撞会消失，所以利用stack的特性来做这题最适合

class Solution {
    public int[] asteroidCollision(int[] asteroids) {
        Stack<Integer> stack = new Stack();

        for (int i = 0; i < asteroids.length; i++) {
            int asteroid = asteroids[i];
            if (stack.isEmpty()) {
                stack.push(asteroid);
            }
            // 两个行星不会发生碰撞
            else if (stack.peek() > 0 && asteroid > 0 || stack.peek() < 0 && asteroid < 0
                     || stack.peek() < 0 && asteroid > 0) {
                stack.push(asteroid);
            }
            // 两个行星发生碰撞，stack顶部的行星size较小，会消失
            else if (Math.abs(stack.peek()) < Math.abs(asteroid)) {
                stack.pop();
                --i;
            }
            // 两个行星发生碰撞，当前的行星size较小，则直接跳过
            else if (Math.abs(stack.peek()) > Math.abs(asteroid)) {
                continue;
            } 
            // 两个size相同的行星发生碰撞，两个都消失
            else {
                stack.pop();
            }
        }

        int size = stack.size();
        int[] res = new int[size];

        while (!stack.isEmpty()) {
            res[--size] = stack.pop();
        }
        return res;
    }
}

3. Time & Space Complexity

时间复杂度和空间复杂度都是O(n)