Description

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

Input: flowerbed = [1,0,0,0,1], n = 1
Output: True

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.

Solution

這道題是個easy題，但是如果沒有想清楚思路就動手寫的話很容易因為沒有考慮全面而越寫越混亂，這也是此題通過率只有百分之三十左右的原因。

由題可知，要放入鮮花的位置兩邊都不能有鮮花，即對於1 0 0 0 1 鮮花只能放在下標為2的位置。但要注意有一個例外，即當頭尾為0的時候：0 0 1 1 0 0 ，在這種情況下鮮花既可以放在第一個位置，也可以放在最後一個位置。

我們每次找到一個可以放入鮮花的位置，就將計數器加一，最終用計數器與傳入的引數n比較大小就可以知道是否能插入鮮花了。

class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
    	int cap = 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)) {
    			cap++;
    			flowerbed[i] = 1;
    		}
    	}
        return cap - n >= 0 ? true : false;
    }
}