阅读 1

LeetCode | 0605. Can Place Flowers种花问题【Python】

LeetCode 0605. Can Place Flowers种花问题【Easy】【Python】【贪心】

Problem

LeetCode

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
复制代码

Example 2:

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

Note:

  1. The input array won't violate no-adjacent-flowers rule.
  2. The input array size is in the range of [1, 20000].
  3. n is a non-negative integer which won't exceed the input array size.

问题

力扣

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
复制代码

示例 2:

输入: flowerbed = [1,0,0,0,1], n = 2
输出: False
复制代码

注意:

  1. 数组内已种好的花不会违反种植规则。
  2. 输入的数组长度范围为 [1, 20000]。
  3. 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
复制代码

代码地址

GitHub链接

关注下面的标签,发现更多相似文章
评论