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.

样例

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

算法1

(枚举) $O(n)$

碰到连续三个0就放一朵花，update花的数量直至为零返回true

时间复杂度

O(n)遍历一次
O(1)

参考文献

C++ 代码

class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        for(int i=0;i<flowerbed.size();i++)
        {
            if(n==0) return true;
            if(flowerbed[i]==0)
            {
                int next=(i==flowerbed.size()-1)?0:flowerbed[i+1];
                int pre=(i==0)?0:flowerbed[i-1];
                if(next+pre==0)
                {
                    flowerbed[i]=1;
                    n--;
                }
            }

        }
        return n<=0;
    }
};