LeetCode Can Place Flowers

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 numbern, 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 = 1Output: True

Example 2:

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

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.

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}};