LeetCode #735 - Asteroid Collision

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本文详细解析了一种行星碰撞模拟算法，该算法通过处理一排行星的数组，根据行星的大小和方向，模拟它们之间的碰撞过程。文章通过多个示例说明了算法的工作原理，包括不同方向和大小的行星相遇时的碰撞结果。

题目描述：

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].

class Solution {
public:
    vector<int> asteroidCollision(vector<int>& asteroids) {
        vector<int> result;
        int i=0;
        while(i<asteroids.size()) //每次遍历到一个元素，就和result的最后一个元素比较
        {
            if(result.size()==0) 
            {
                result.push_back(asteroids[i]);
                i++;
            }
            else 
            {   // result末尾元素向右，且新元素向左才可能碰撞
                if(result.back()>0&&asteroids[i]<0) 
                {
                    // 新元素绝对值大于result的最后一个元素，那么删除result的最后一个元素，保持i不变进入下一个循环
                    if(abs(asteroids[i])>abs(result.back())) result.pop_back();
                    else if(abs(asteroids[i])<abs(result.back())) i++;
                    else
                    {
                        result.pop_back();
                        i++;
                    }
                    
                }
                else //不会碰撞的情况下，将新元素加入
                {   
                    result.push_back(asteroids[i]);
                    i++;
                }
            }
        }
        return result;
    }
};