【leetcode】1375. Bulb Switcher III

题目如下：

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]

解题思路：用两个列表分别保存所有开着的灯的下标和所有关着的灯的下标，每次只要判断最大的亮着的灯下标是否小于最小的关着的灯的下标即可。

代码如下：

class Solution(object):
    def numTimesAllBlue(self, light):
        """
        :type light: List[int]
        :rtype: int
        """
        import bisect
        on = []
        off = range(1,len(light)+1)

        res = 0
        for i in range(len(light)):
            inx = bisect.bisect_left(off,light[i])
            del off[inx]
            bisect.insort_left(on,light[i])

            if len(off) == 0 or on[-1] < off[0]:
                res += 1
        return res