[LeetCode]Asteroid Collision

题目描述:

LeetCode 735. Asteroid Collision

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].

题目大意:

给定一组整数表示一排小行星的体积,正数表示向右运动,负数表示向左运动。

运动方向相反的小行星会发生碰撞,体积小的小行星会被撞毁,若体积相同,则“同归于尽”。

求碰撞结束后的小行星列表。

解题思路:

栈(Stack)

初始化空栈nlist,用来保存所有互不冲突的小行星

遍历asteroids,记当前小行星为n

当栈非空,栈顶元素向右运动,n向左运动后可以将栈顶撞毁时,重复弹出栈

如果栈为空,或者栈顶元素向左运动,或者n向右运动时,将n入栈nlist

否则,如果n向左运动,并且与栈顶元素同归于尽时,将nlist栈顶弹出

Python代码:

class Solution(object):
    def asteroidCollision(self, asteroids):
        """
        :type asteroids: List[int]
        :rtype: List[int]
        """
        nlist = []
        for n in asteroids:
            while nlist and nlist[-1] > 0 and nlist[-1] + n < 0:
                nlist.pop()
            if not nlist or nlist[-1] < 0 or n > 0:
                nlist.append(n)
            elif n < 0 and nlist[-1] + n == 0:
                nlist.pop()
        return nlist

 

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