【python3】leetcode 605. Can Place Flowers（easy）

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605. Can Place Flowers（easy）

Suppose you have a long flowerbed in which some of the plots are planted and some are not. However, flowers cannot be planted in adjacent plots - they would compete for water and both would die.

Given a flowerbed (represented as an array containing 0 and 1, where 0 means empty and 1 means not empty), and a number n, return if n new flowers can be planted in it without violating the no-adjacent-flowers rule.

Example 1:

Input: flowerbed = [1,0,0,0,1], n = 1
Output: True

Example 2:

Input: flowerbed = [1,0,0,0,1], n = 2
Output: False

题目：间隔插入n个花

1 stupid me 

感觉自己越来越啰嗦了Or2,

先把几个边界或简单的特殊插入判断返回：

       n = 0 ->true

      len(flowerbed) == 1 & flowerbed[0] = 0 -> true

      由于是间隔插入 所以如果 n>(len)/2 + 1也不能插完 -》false

剩下的就是普通的插入了，遍历整个列表，如果条件满足就插入 同时 n-1

class Solution:
    def canPlaceFlowers(self, flowerbed, n):
        """
        :type flowerbed: List[int]
        :type n: int
        :rtype: bool
        """
        length = len(flowerbed)
        if (length == 1 and flowerbed[0] == 0 and n ==1) or n == 0:return True
        if n > int(length/2) + 1:False
        i = 0
        while(i < length and n > 0):
            flag = 0
            if flowerbed[i] == 0:
                if i == 0 and flowerbed[1] == 0:flag = 1
                elif i == length - 1 and flowerbed[i-1] == 0:flag = 1
                elif flowerbed[i-1]!=1 and flowerbed[i+1] != 1:flag = 1
            if flag == 1:
                flowerbed[i] = 1
                n -= 1  
            i += 1
        return n == 0

Runtime: 120 ms, faster than 6.13% of Python3

或者把所有的判断写在一起 ，我觉得不是很直观

class Solution:
    def canPlaceFlowers(self, flowerbed, n):
        """
        :type flowerbed: List[int]
        :type n: int
        :rtype: bool
        """
        length = len(flowerbed)
        if (length == 1 and flowerbed[0] == 0 and n ==1) or n == 0:return True
        if n > int(length/2) + 1:False
        i = 0
        while(i < length and n > 0):
            if flowerbed[i] == 0 and ((i == 0 and flowerbed[1] == 0) or ((i == length - 1) and flowerbed[i - 1] == 0) or (i < length - 1 and i > 0 and flowerbed[i+1] == 0 and flowerbed[i-1] == 0)):
                flowerbed[i] = 1
                n -= 1  
            i += 1
        return n == 0