题目地址：https://leetcode.com/problems/can-place-flowers/description/
题目描述
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、 Theinputarraywon'tviolateno-adjacent-flowersrule.；
2、 Theinputarraysizeisintherangeof[1,20000].；
3、 nisanon-negativeintegerwhichwon'texceedtheinputarraysize.；
解题方法
贪婪算法
这个题做的方式很蠢，就是一次遍历，看每个位置能不能种花，如果可以就种上，否则就判断下一个节点。
不能种花的条件是：
1、 已经有花；
2、 当i>0时，右边有花；
3、 当i<len-1时，左边有花；
注意两点：
1、 遍历的时候如果该位置能种花，则种上，否则会影响下一个位置的判断；
2、 最后的条件是n<=0，即能种花的位置比给出的n多；
class Solution(object):
def canPlaceFlowers(self, flowerbed, n):
"""
:type flowerbed: List[int]
:type n: int
:rtype: bool
"""
for i, num in enumerate(flowerbed):
if num == 1: continue
if i > 0 and flowerbed[i - 1] == 1: continue
if i < len(flowerbed) - 1 and flowerbed[i + 1] == 1: continue
flowerbed[i] = 1
n -= 1
return n <= 0
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二刷，做法思路一样：有连续的三个0的话，中间的这个位置种上一朵花。
python代码：
class Solution(object):
def canPlaceFlowers(self, flowerbed, n):
"""
:type flowerbed: List[int]
:type n: int
:rtype: bool
"""
flowerbed = [0] + flowerbed + [0]
N = len(flowerbed)
res = 0
for i in range(1, N - 1):
if flowerbed[i - 1] == flowerbed[i] == flowerbed[i + 1] == 0:
res += 1
flowerbed[i] = 1
return res >= n
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C++代码如下：
class Solution {
public:
bool canPlaceFlowers(vector<int>& flowerbed, int n) {
flowerbed.insert(flowerbed.begin(), 0);
flowerbed.push_back(0);
int N = flowerbed.size();
int res = 0;
for (int i = 1; i < N - 1; ++i) {
if (flowerbed[i - 1] == 0 && flowerbed[i] == 0 && flowerbed[i + 1] == 0) {
++res;
flowerbed[i] = 1;
}
}
return res >= n;
}
};
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