leeetcode 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]..

一开始没有认真思考，直接用一个队列去存储正数，然后遇到负数就进行合并判断。时间复杂度接近O(n^{2)但一定小于n}2.下面是代码。其实可以用s代替ans见第二分代码.

class Solution {
public:
    vector<int> asteroidCollision(vector<int>& asteroids) {
        vector<int> s;
        vector<int> ans;
        for (auto x : asteroids) {
            if (x > 0) s.emplace_back(x);
            else {
                int mark = 0;
                while (!s.empty()) {
                    int u = s.back();
                    if (-x > u) s.pop_back();
                    else if (-x == u) {
                        mark = 1;
                        s.pop_back(); break;
                    } else {
                        break;
                    }
                }
                if (s.empty() && !mark) ans.push_back(x);
            }
        }
        for (auto x : s) {
            ans.push_back(x);
        }
        return ans;
    }
};

用一个s就可以解决，如果s中有负数那么就不用在往前找了(前面不可能有正数了).

优化后的代码

class Solution {
public:
    vector<int> asteroidCollision(vector<int>& asteroids) {
        vector<int> s;
        for (auto x : asteroids) {
            if (x > 0) s.emplace_back(x);
            else if (x < 0) {
                int y = -x;
                int mark = 1;
                while (!s.empty() && s.back() > 0) {
                    if (s.back() >= y) {
                        if (s.back() == y) s.pop_back();
                        mark = 0;
                        break;
                    } else {
                        s.pop_back();
                    }
                }
                if (mark) s.emplace_back(x);
            }
        }
        return s;
    }
};