> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/605.can_place_flowers.md).

# 605.Can Place Flowers

**605.Can Place Flowers**

难度:Easy

> 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.
```

可以创造一个以10开始的并以01结尾的数组，然后比较每两个1之间的间隔，可以放几个flower。比较其与n的大小。\
例子中，假象start的索引为-2

```
class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        flowerbed.push_back(0);
        flowerbed.push_back(1);
        int total=0;
        int start=-2;
        for(int i=0; i<flowerbed.size();i++ )
        {
            if(flowerbed[i] ) {
                int r=i-start-3;
                 start=i;
            //    cout<<r<<endl;
                if(r<1 ) continue;

                total += (r+ r%2) /2;
               
            } 
        }
      //  cout<<total<<endl;
        return total>=n;
    }
};
```

> 执行用时 :20 ms, 在所有 C++ 提交中击败了79.42%的用户\
> 内存消耗 :11.2 MB, 在所有 C++ 提交中击败了17.02%的用户


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