LeetCode | 1375. Bulb Switcher III灯泡开关 III【Python】

LeetCode 1375. Bulb Switcher III灯泡开关 III【Medium】【Python】【最大编号】

Problem

LeetCode

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]

问题

力扣

房间中有 n 枚灯泡,编号从 1 到 n,自左向右排成一排。最初,所有的灯都是关着的。

在 k 时刻( k 的取值范围是 0 到 n - 1),我们打开 light[k] 这个灯。

灯的颜色要想 变成蓝色 就必须同时满足下面两个条件:

  • 灯处于打开状态。
  • 排在它之前(左侧)的所有灯也都处于打开状态。

请返回能够让 所有开着的 灯都 变成蓝色 的时刻 数目

示例 1:

输入:light = [2,1,3,5,4]
输出:3
解释:所有开着的灯都变蓝的时刻分别是 1,2 和 4 。

示例 2:

输入:light = [3,2,4,1,5]
输出:2
解释:所有开着的灯都变蓝的时刻分别是 3 和 4(index-0)。

示例 3:

输入:light = [4,1,2,3]
输出:1
解释:所有开着的灯都变蓝的时刻是 3(index-0)。
第 4 个灯在时刻 3 变蓝。

示例 4:

输入:light = [2,1,4,3,6,5]
输出:3

示例 5:

输入:light = [1,2,3,4,5,6]
输出:6

提示:

  • n == light.length
  • 1 <= n <= 5 * 10^4
  • light[1, 2, ..., n] 的一个排列。

思路

最大编号

遍历一遍,修改当前状态为开灯,并比较开灯次数是否等于记录的最大编号。
如果相等,表示灯可以变蓝,于是就计数 + 1。

时间复杂度: O(n)
空间复杂度: O(1)

Python3代码
class Solution:
    def numTimesAllBlue(self, light: List[int]) -> int:
        n = len(light)
        flag = [False] * (n+1)
        cnt_blue, cnt_light = 0, 0
        max_i = 0
        for i in range(n):
            flag[light[i]] = True  # turn on bulb开灯
            cnt_light += 1
            max_i = max(max_i, light[i])
            # 已开灯数量 = 当前记录的最大编号:灯变蓝计数 + 1
            if cnt_light == max_i:
                cnt_blue += 1
        return cnt_blue

代码地址

GitHub链接

posted @ 2020-03-08 15:51  Wonz  阅读(244)  评论(0编辑  收藏  举报