LeetCode Can Place Flowers

最新推荐文章于 2025-08-09 17:01:50 发布

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最新推荐文章于 2025-08-09 17:01:50 发布

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分类专栏：算法文章标签： leetcode

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解决一个经典的编程问题：如何在不违反相邻花朵不能种植的规则下，在给定的花坛中种植尽可能多的新花。通过一次遍历数组，判断是否可以在空位种植花朵，并考虑边界情况。

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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.

本题思路比较简单，循环一遍，该位置左右都为0则将其赋值为1，注意边界点的特殊情况。

Code：

class Solution {
public:
	bool canPlaceFlowers(vector<int>& flowerbed, int n) {
		if (n == 0)return true;
		//flowerbed为空
		if (flowerbed.size() == 0)return false;
		//flowerbed只有一个坑
		if (flowerbed.size() == 1)
		{
			return flowerbed[0] == 0;
		}
		//处理两个特殊的点，即两个边界点
		if (flowerbed[0] == 0 && flowerbed[1] == 0) {
			flowerbed[0] = 1;
			n--;
		}

		for (int i = 0; i < flowerbed.size() - 1; i++) {
			if (flowerbed[i] == 1) {
				continue;
			}
			if (flowerbed[i - 1] == 0 && flowerbed[i + 1] == 0)
			{
				flowerbed[i] = 1;
				n--;
			}
		}
		//最后一个点
		if (flowerbed[flowerbed.size() - 2] == 0 && flowerbed[flowerbed.size() - 1] == 0){
			flowerbed[flowerbed.size() - 1] = 1;
			n--;
		}
		return n <= 0;//小于等于0表示至少能种n棵树
	}
};