LeetCode 605. Can Place Flowers （可以种花）

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.

题目标签：Array

　　题目给了我们一个 flowerbed array， 和一个 n， 让我们找出是否右足够的空间来置放 n 个花。 每一朵符合条件的花，它的左右相邻位置 不能有花。

　　所以我们要遍历flowerbed array：

　　　　如果遇到0的话，还要检查这个0的左右两边是否都是0，都是0的情况下，才可以算作一个空间来置放花，然后可以额外跳过右边的0；

　　　　如果遇到1的话，只需要额外跳过1右边的位置；

　　　　一旦当空间数量 已经 等于 n 了，就不需要继续找下去了，直接返回true。（这里要加一个 = ，因为给的 n 有可能是 0）

　　当遍历完了，说明空间不够，返回 false 即可。

Java Solution:

Runtime beats 60.45% 

完成日期：10/15/2017

关键词：Array

关键点：符合条件的empty spot 的 左右邻居也要为0

class Solution
{
public boolean canPlaceFlowers(int[] flowerbed, int n)
{
int count = 0; // count the valid empty spot

for(int i=0; i<flowerbed.length; i++)
{

if(flowerbed[i] == 0) // if find empty spot, check its adjacent spots
{
if(i-1 >= 0 && flowerbed[i-1] == 1) // check left adjacent spot
continue;
if(i+1 < flowerbed.length && flowerbed[i+1] == 1) // check right adjacent spot
continue;

// if come here, meaning this 0 is valid spot to place flower
count++;
i++; // skip next 0
}
else if(flowerbed[i] == 1) // if find a flower spot, its next spot is not valid, skip it
{
i++;
}

if(count >= n) // if there are enough spots to place flower, no need to continue
return true;

}

return false;
}
}

参考资料：N/A

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