LeetCode 0605. Can Place Flowers种花问题【Easy】【Python】【贪心】
Problem
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
Note:
- The input array won't violate no-adjacent-flowers rule.
- The input array size is in the range of [1, 20000].
- n is a non-negative integer which won't exceed the input array size.
问题
假设你有一个很长的花坛，一部分地块种植了花，另一部分却没有。可是，花卉不能种植在相邻的地块上，它们会争夺水源，两者都会死去。
给定一个花坛（表示为一个数组包含0和1，其中0表示没种植花，1表示种植了花），和一个数 n 。能否在不打破种植规则的情况下种入 n 朵花？能则返回True，不能则返回False。
示例 1:
输入: flowerbed = [1,0,0,0,1], n = 1
输出: True
示例 2:
输入: flowerbed = [1,0,0,0,1], n = 2
输出: False
注意:
- 数组内已种好的花不会违反种植规则。
- 输入的数组长度范围为 [1, 20000]。
- n 是非负整数，且不会超过输入数组的大小。
思路
贪心
从反面来看，不能种花的情况共有三种：
- 当前位置已有花
- 位置不是头，且左边已有花
- 位置不是尾，且右边已有花
其余情况可以种花。
时间复杂度: O(len(flowerbed)) 空间复杂度: O(1)
Python代码
class Solution(object):
def canPlaceFlowers(self, flowerbed, n):
"""
:type flowerbed: List[int]
:type n: int
:rtype: bool
"""
if not flowerbed or len(flowerbed) == 0:
return False
cnt = 0
for plot in range(len(flowerbed)): # range:0 - len(flowerbed)-1
if flowerbed[plot] == 1: # already planted flower
continue
if plot > 0 and flowerbed[plot-1] == 1: # left planted flower
continue
if plot < len(flowerbed) - 1 and flowerbed[plot + 1] == 1: # right planted flower
continue
flowerbed[plot] = 1
cnt += 1
if cnt >= n:
return True
else:
return False
