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:
代码语言：javascript复制Input: flowerbed = [1,0,0,0,1], n = 1
Output: TrueExample 2:
代码语言：javascript复制Input: flowerbed = [1,0,0,0,1], n = 2
Output: FalseNote:
- 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.
我的答案:
代码语言：javascript复制class Solution {
public boolean canPlaceFlowers(int[] flowerbed, int n) {
List<Integer> list = new ArrayList<>();
for (int i = 0; i < flowerbed.length; i ) {
if(flowerbed[i] == 1){
list.add(i);
}
}
if(list.size()==0 ){
n = n-(flowerbed.length 1)/2;
return n<=0;
}
//head
Integer head = list.get(0);
if(head>1){
n = n-head/2;
}
//tail
Integer tail = list.get(list.size()-1);
if(flowerbed.length-tail-1>1){
n = n-(flowerbed.length-tail-1)/2;
}
for(int i = 0;i<list.size()-1;i ){
int begin = list.get(i);
int end = list.get(i 1);
if(end-begin>2){
n = n-(end-begin-2)/2;
}
}
return n <= 0;
}
}大牛的答案:
代码语言：javascript复制public boolean canPlaceFlowers(int[] flowerbed, int n) {
int count = 1;
int result = 0;
for(int i=0; i<flowerbed.length; i ) {
if(flowerbed[i] == 0) {
count ;
}else {
result = (count-1)/2;
count = 0;
}
}
if(count != 0) result = count/2;
return result>=n;
}
