LeetCode每日一题(Bulb Switcher III)

最新推荐文章于 2022-03-31 09:58:37 发布

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这篇博客介绍了如何利用二叉堆解决一个关于灯泡颜色变化的问题。给定一个灯泡数组，每个灯泡会在特定时刻变亮，当一个灯泡及其左侧所有灯泡都亮起时，它会变为蓝色。博客内容包含若干示例输入和输出，展示了在不同情况下灯泡全蓝的时刻数量。博主提出使用二叉堆来跟踪灯泡状态，并给出了用Rust实现的代码示例。

There is a room with n bulbs, numbered from 1 to n, arranged in a row from left to right. Initially, all the bulbs are turned off.

At moment k (for k from 0 to n - 1), we turn on the light[k] bulb. A bulb change color to blue only if it is on and all the previous bulbs (to the left) are turned on too.

Return the number of moments in which all turned on bulbs are blue.

Example 1:

Input: light = [2,1,3,5,4]
Output: 3

Explanation: All bulbs turned on, are blue at the moment 1, 2 and 4.

Example 2:

Input: light = [3,2,4,1,5]
Output: 2

Explanation: All bulbs turned on, are blue at the moment 3, and 4 (index-0).

Example 3:

Input: light = [4,1,2,3]
Output: 1

Explanation: All bulbs turned on, are blue at the moment 3 (index-0).
Bulb 4th changes to blue at the moment 3.

Example 4:

Input: light = [2,1,4,3,6,5]
Output: 3

Example 5:

Input: light = [1,2,3,4,5,6]
Output: 6

Constraints:

n == light.length
1 <= n <= 5 * 10^4
light is a permutation of [1, 2, …, n]

建一个binary heap, 将数组里的数一个个的放到heap中，当heap中的最大值与heap的长度相等的时候所有灯泡都是蓝色的。

代码实现(Rust):


use std::collections::BinaryHeap;

impl Solution {
    pub fn num_times_all_blue(light: Vec<i32>) -> i32 {
         let mut heap: BinaryHeap<i32> = BinaryHeap::new();
         let mut ans = 0;
         for l in light {
             heap.push(l);
             if *heap.peek().unwrap() as usize == heap.len() {
                 ans += 1;
             }
         }
         ans
    }
}