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:
- 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.
这道题采用贪心算法,满足连续三个零就是一个符合点,一次遍历即可,首尾的边界问题(只需要两个连续零)不能忽略。
C++实现如下:
bool canPlaceFlowers(vector<int>& flowerbed, int n) {
int length=flowerbed.size();
int count=0;
for(int i=0;i<length;)
{
while(flowerbed[i]==0)
{
i++;
if((i==length-1||i==1)&&flowerbed[i]==0)
count++;
else if(i<length-1&&flowerbed[i]==0&&flowerbed[++i]==0)
count++;
}
i++;
if(n<=count)
return true;
}
return false;
}