题目描述：

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 = 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.

 

要完成的函数：

bool canPlaceFlowers(vector<int>& flowerbed, int n) 

 

说明：

1、假设你是一个花农，要在花床上种花，花不能相邻种植，不然会彼此竞争空气、水分和营养物质。现在给你一个vector和一个整数n，vector中的0代表此处没有种花，1代表此处种花，要求判断能不能在花床上再种下n株花，彼此不能相邻。

2、这道题目不难，先对最开始的边界情况做处理，接着再做一次遍历直到倒数第二位，最后再对最后一个数做边界处理。

中间处理的过程是逐个判断，对应不同的数值做不同的操作。

代码如下：（附详解）

bool canPlaceFlowers(vector
     
      & flowerbed, int n)     {        int s1=flowerbed.size(),i=0,count=0;        if(flowerbed[i]==1)//如果最开始是1，那么得从第2位（从0开始）开始考虑            i+=2;        else//如果最开始是0        {            if(flowerbed[i+1]==0)//如果下一位还是0，那么count++，下一次从i=2开始处理            {                count++;                i+=2;            }            else//如果下一位是1，那么下一次就得从i=3开始考虑                i+=3;        }        while(i
     

上述代码实测18ms，beats 99.20% of cpp submissions。