735. Asteroid Collision彗星相撞后的消失数组

［抄题］：

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.

 [暴力解法]：

时间分析：

空间分析：

 [优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思维问题］：

不知道怎么改数组，忘了：新建一个linked list之后动态添加即可

［英文数据结构或算法，为什么不用别的数据结构或算法］：

s.stream().mapToInt(i->i).toArray(); integer对象转为数字流，然后再转数组。

［一句话思路］：

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

1. 不理解：减少一位需要i--，绝对值相等时+1-1，不用i--

［二刷］：

［三刷］：

［四刷］：

［五刷］：

  [五分钟肉眼debug的结果]：

［总结］：

对linkedlist可以用polllast getlast的方法

［复杂度］：Time complexity: O(n) Space complexity: O(n)

［算法思想：迭代/递归/分治/贪心］：

［关键模板化代码］：

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

 [代码风格] ：

 [是否头一次写此类driver funcion的代码] ：

 [潜台词] ：

 

class Solution {
    public int[] asteroidCollision(int[] asteroids) {
        //initialization: array
        LinkedList<Integer> list = new LinkedList<>();
        
        //for loop: add to array if positive, or compare and poll
        for (int i = 0; i < asteroids.length; i++) {
            if (asteroids[i] > 0 || list.size() == 0 || list.getLast() < 0) {
                list.add(asteroids[i]);
            }else {
                //if bigger or equal, i--
                if (list.getLast() <= -asteroids[i]) {
                    //if bigger, romove only the last
                    if (list.pollLast() < -asteroids[i]) {
                        i--;
                    }
                }
            }
        }
        
        //return .stream.to.toarray
        return list.stream().mapToInt(i -> i).toArray();
    }
}

View Code

 

转载于:https://www.cnblogs.com/immiao0319/p/9509987.html