605. Can Place Flowers【easy】

 

605. Can Place Flowers【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:

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.

 

错误解法：

class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        int max = 0;
        int temp = 0;
        
        for (int i = 0; i < flowerbed.size(); ++i)
        {
            if (flowerbed[i] == 0)
            {
                temp++;
                max = temp > max ? temp : max;
            }
            else
            {
                temp = 0;
            }
        }

        
        if ((max - 2) % 2)
        {
            return ((max - 2) / 2 + 1 >= n); 
        }
        else
        {
            return ((max - 2) / 2 >= n);
        }
    }
};

想要用最长连续的0去求，调试了很久，但还是条件判断有问题，这种方法不妥！

 

参考大神们的解法，如下：

解法一：

public class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
        int count = 0;
        for(int i = 0; i < flowerbed.length && count < n; i++) {
            if(flowerbed[i] == 0) {
         //get next and prev flower bed slot values. If i lies at the ends the next and prev are considered as 0. 
               int next = (i == flowerbed.length - 1) ? 0 : flowerbed[i + 1]; 
               int prev = (i == 0) ? 0 : flowerbed[i - 1];
               if(next == 0 && prev == 0) {
                   flowerbed[i] = 1;
                   count++;
               }
            }
        }
        
        return count == n;
    }
}

 

解法二：

public boolean canPlaceFlowers(int[] flowerbed, int n) {
    for (int idx = 0; idx < flowerbed.length && n > 0; idx ++)
        if (flowerbed [idx] == 0 && (idx == 0 || flowerbed [idx - 1] == 0) && (idx == flowerbed.length - 1 || flowerbed [idx + 1] == 0)) {
            n--;
            flowerbed [idx] = 1;
        }
    return n == 0;
}

解法一和解法二都需要随时更新原来数组的状态

 

解法三：

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

不更新原来数组的值，通过多移下标来实现