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.

第一种：3格的滑动窗口

class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        int str = 0 ,end = 2;
        int tol = 0;
        if (flowerbed.size()==1&&flowerbed[0]==0) {
            tol++;
        }else if(flowerbed[0]==0&&flowerbed[1]==0) {
            flowerbed[0]=1;
            tol++;
        }
        
        while (end<=flowerbed.size()-1) {
            if (!(flowerbed[str]||flowerbed[str+1]||flowerbed[end])) {
                flowerbed[str+1] = 1;
                tol++;
            }
            str++;
            end++;
        }
        if (flowerbed.size()>=3&&flowerbed[flowerbed.size()-1]==0&&flowerbed[flowerbed.size()-2]==0) {
            tol++;
        }
        return tol>=n?1:0;
    }
};

第二种：

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