leetcode605- 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.

 

算法：贪婪法。一个count记录当前种了多少了。for循环遍历一次，看到0的，而且前面是0或者头，而且后面是0或者尾，就可以种下去并加计数器。

细节：小心边界条件，比如就种0个花的。跑一遍改结构。

 

实现：

class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
        if (flowerbed == null || flowerbed.length == 0) {
            return false;
        }
        int count = 0;
        for (int i = 0; i < flowerbed.length; i++) {
            if (flowerbed[i] == 0
                && (i == 0 || flowerbed[i - 1] == 0) 
                && (i == flowerbed.length - 1 || flowerbed[i + 1] == 0)) {
                count ++;
                // 改变传入的数据，需要问是否允许
                flowerbed[i] = 1;
            }
            if (count >= n) {
                    return true;
            }
        }
        return false;
    }
}