小行星碰撞

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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].

类似于计算四则运算，区别在于此处的优先级为碰撞，如果有碰撞则优先计算，因此使用栈的思想，碰撞时弹出消除，最后栈的内容就是最终结果。

解法：

vector<int> asteroidCollision(vector<int>& a) {
    vector<int> s; // use vector to simulate stack.
    for (int i = 0; i < a.size(); i++) {
        if (a[i] > 0 || s.empty() || s.back() < 0) // a[i] is positive star or a[i] is negative star and there is no positive on stack
            s.push_back(a[i]);
        else if (s.back() <= -a[i]) { // a[i] is negative star and stack top is positive star
            if(s.back() < -a[i]) i--; // only positive star on stack top get destroyed, stay on i to check more on stack.
            s.pop_back(); // destroy positive star on the frontier;
        } // else : positive on stack bigger, negative star destroyed.
    }
    return s;
}