LeetCode-Can Place Flowers

最新推荐文章于 2024-04-04 23:56:46 发布

原创最新推荐文章于 2024-04-04 23:56:46 发布 · 231 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LeetCode #Java

LeetCode 专栏收录该内容

306 篇文章

订阅专栏

解决一个特定的问题，即在一个只包含0和1的一维数组中，0表示空位置，1表示已种植花朵的位置。需要判断能否在遵守不相邻种花规则的前提下种植n朵新花。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description:
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.

题意：给定一个只包含0和1的一位数组，0表示这个位置是空的，1表示这个位置上已经种植了花；现在给定还需要种植的花数n,要求两朵花之间的最小距离为2，即种植花的位置的左右相邻位置均不能够种植花，判断是否还能种植给定花数；

解法：我们只需要在遍历数组的时候，每当遇到一个0，就判断其是否可以种植，满足左右相邻位置均为空即可；若能够种植，将此位置置为1，并且将n值减1，否则继续寻找下一个0，在遍历的过程中，当n值减为1时则返回true；

class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
        if (n == 0) {
            return true;
        }
        for (int i = 0; i < flowerbed.length; i++) {
            if (flowerbed[i] == 0 && (i == 0 || flowerbed[i - 1] != 1) &&
               (i  == flowerbed.length - 1 || flowerbed[i + 1] != 1)) {
                n--;
                flowerbed[i] = 1;
            }
            if (n == 0) {
                return true;
            }
        }
        return false;
    }
}