转自https://blog.csdn.net/xiakexiaohu/article/details/78639927
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]..
分析：
class Solution {
public int[] asteroidCollision(int[] as) {
//给定一堆数字模拟小行星碰撞（正数代表向右，负数代表向左）,元素代表大小。当两个行星碰撞时，小的会爆炸(类似碰碰球)。
//如果尺寸大小相同，则都会爆炸。相同方向的永远不会碰撞。返回其碰撞的状态。
//思路：利用stack存储，每次对比栈顶元素（相同的话就放入栈中），否则就对比，直至栈顶相同为止
//注意：当栈顶为负数时，直接进栈。只有当栈顶为正数才需要发生碰撞
Stack<Integer> stack=new Stack<Integer>();
for(int i=0;i<as.length;i++){
//如果栈为空或者大于0
if(stack.isEmpty()||as[i]>0){
stack.push(as[i]);
continue;
}
//遇见不同方向元素
while(true){
int pre=stack.peek();
if(pre<0){
//当栈顶为负数时
stack.push(as[i]);
break;
}else if(pre==-as[i]){
//两个元素相同,消掉栈顶元素
stack.pop();
break;
}else if(pre>-as[i]){
//栈顶元素更大，消掉
break;
}else {
//栈顶元素更小，依次迭代和栈顶比较，直至满足条件
stack.pop();
//如果栈为空了，就直接放入，否则继续循环比较
if(stack.isEmpty()){
stack.push(as[i]);
break;
}
}
}
}
//将stack结果输出
int [] res=new int[stack.size()];
int i=stack.size()-1;
while(!stack.empty()){
res[i--]=stack.pop();
}
return res;
}
}