LeetCode每日一题(2251. Number of Flowers in Full Bloom)

You are given a 0-indexed 2D integer array flowers, where flowers[i] = [starti, endi] means the ith flower will be in full bloom from starti to endi (inclusive). You are also given a 0-indexed integer array persons of size n, where persons[i] is the time that the ith person will arrive to see the flowers.

Return an integer array answer of size n, where answer[i] is the number of flowers that are in full bloom when the ith person arrives.

Example 1:

Input: flowers = [[1,6],[3,7],[9,12],[4,13]], persons = [2,3,7,11]
Output: [1,2,2,2]

Explanation: The figure above shows the times when the flowers are in full bloom and when the people arrive.
For each person, we return the number of flowers in full bloom during their arrival.

Example 2:

Input: flowers = [[1,10],[3,3]], persons = [3,3,2]
Output: [2,2,1]

Explanation: The figure above shows the times when the flowers are in full bloom and when the people arrive.
For each person, we return the number of flowers in full bloom during their arrival.

Constraints:

• 1 <= flowers.length <= 5 * 104
• flowers[i].length == 2
• 1 <= starti <= endi <= 109
• 1 <= persons.length <= 5 * 104
• 1 <= persons[i] <= 109

---

1. 将 flowers 按开始时间(start)进行排序, 如果开始时间(start)相同则按结束时间(end)排序
2. 用一个 min heap 来维护当前开放的花， heap 中保存的是花的凋谢时间
3. 按时间顺序对 persons 进行排序， 使得 persons[i] = (access_time, index)
4. 遍历 persons, 每次遍历将 flowers 中 start <= access_time 的 flower 放到 heap 中，如果遇到 start > access_time 的情况则终止向 heap 中添加 flower 。然后再将 heap 中 end < access_time 的 flower 移出 heap, 此时 heap 中剩余的就是该 person 访问时盛开的 flowers。

---

use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
    pub fn full_bloom_flowers(flowers: Vec<Vec<i32>>, persons: Vec<i32>) -> Vec<i32> {
        let mut flowers: Vec<(i32, i32)> = flowers.into_iter().map(|l| (l[0], l[1])).collect();
        flowers.sort();
        flowers.reverse();
        let mut persons: Vec<(i32, usize)> = persons
            .into_iter()
            .enumerate()
            .map(|(i, v)| (v, i))
            .collect();
        persons.sort();
        let mut ans = vec![0; persons.len()];
        let mut bloomings: BinaryHeap<Reverse<i32>> = BinaryHeap::new();
        for (t, i) in persons {
            while let Some((start, end)) = flowers.pop() {
                if start > t {
                    flowers.push((start, end));
                    break;
                }
                bloomings.push(Reverse(end));
            }
            while let Some(Reverse(end)) = bloomings.pop() {
                if end >= t {
                    bloomings.push(Reverse(end));
                    break;
                }
            }
            ans[i] = bloomings.len() as i32;
        }
        ans
    }
}