0735. Asteroid Collision

# 735. Asteroid Collision#

## 题目 #

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]..

## 解题思路 #

1. 所有向左飞的行星都向左，所有向右飞的行星都向右。
2. 向左飞的行星，如果飞行中没有向右飞行的行星，那么它将安全穿过。
3. 跟踪所有向右移动到右侧的行星，最右边的一个将是第一个面对向左飞行行星碰撞的。
4. 如果它幸存下来，继续前进，否则，任何之前的向右的行星都会被逐一被暴露出来碰撞。

## 代码 #

``````
package leetcode

func asteroidCollision(asteroids []int) []int {
res := []int{}
for _, v := range asteroids {
for len(res) != 0 && res[len(res)-1] > 0 && res[len(res)-1] < -v {
res = res[:len(res)-1]
}
if len(res) == 0 || v > 0 || res[len(res)-1] < 0 {
res = append(res, v)
} else if v < 0 && res[len(res)-1] == -v {
res = res[:len(res)-1]
}
}
return res
}

``````

Sep 6, 2020