Leetcode 605(Java)
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.
除了開頭和結尾的特殊情況外,在中間要種花,都需要有三個空位。因此設定變數empty來計數空位,當達到3時,可以插入一朵花,並且empty變為1。而在開頭和結尾處,則需要兩個空位就能放花,注意[0,0,0,0,1],這一類情況。AC碼:
public class Solution {
public boolean canPlaceFlowers(int[] flowerbed, int n) {
if(n==1&&flowerbed.length==1&&flowerbed[0]==0)return true;
int empty=0 ;
int canflower=0;
for(int i=0;i<flowerbed.length;i++){
if(flowerbed[i]==0){
empty++;
if(empty==3){
canflower++;
empty=1;
}
if(((i==flowerbed.length-1)||(i==1))&&empty>=2){
canflower++;
empty=1;
}
}else{
empty=0;
}
}
if(canflower>=n)return true;
return false;
}
}