LeetCode每日一题(2126. Destroying Asteroids)

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给定一个表示行星质量的整数和一个表示小行星质量的数组，行星可以按任意顺序与小行星碰撞。如果行星的质量大于或等于小行星，小行星被摧毁，行星增加小行星的质量。返回行星能否摧毁所有小行星。示例展示了一种判断方法：先对小行星数组排序，然后从小到大依次碰撞。

You are given an integer mass, which represents the original mass of a planet. You are further given an integer array asteroids, where asteroids[i] is the mass of the ith asteroid.

You can arrange for the planet to collide with the asteroids in any arbitrary order. If the mass of the planet is greater than or equal to the mass of the asteroid, the asteroid is destroyed and the planet gains the mass of the asteroid. Otherwise, the planet is destroyed.

Return true if all asteroids can be destroyed. Otherwise, return false.

Example 1:

Input: mass = 10, asteroids = [3,9,19,5,21]
Output: true

Explanation: One way to order the asteroids is [9,19,5,3,21]:

The planet collides with the asteroid with a mass of 9. New planet mass: 10 + 9 = 19
The planet collides with the asteroid with a mass of 19. New planet mass: 19 + 19 = 38
The planet collides with the asteroid with a mass of 5. New planet mass: 38 + 5 = 43
The planet collides with the asteroid with a mass of 3. New planet mass: 43 + 3 = 46
The planet collides with the asteroid with a mass of 21. New planet mass: 46 + 21 = 67

All asteroids are destroyed.

Example 2:

Input: mass = 5, asteroids = [4,9,23,4]
Output: false

Explanation:
The planet cannot ever gain enough mass to destroy the asteroid with a mass of 23.
After the planet destroys the other asteroids, it will have a mass of 5 + 4 + 9 + 4 = 22.
This is less than 23, so a collision would not destroy the last asteroid.

Constraints:

1 <= mass <= 10^5
1 <= asteroids.length <= 10^5
1 <= asteroids[i] <= 10^5

将 asteroids 排序，然后从小到大碰过去就好了

impl Solution {
    pub fn asteroids_destroyed(mass: i32, mut asteroids: Vec<i32>) -> bool {
        let mut mass = mass as i64;
        asteroids.sort();
        for a in asteroids {
            if a as i64 > mass {
                return false;
            }
            mass += a as i64;
        }
        true
    }
}