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.
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
1 class Solution 2 { 3 public boolean canPlaceFlowers(int[] flowerbed, int n) 4 { 5 int count = 0; // count the valid empty spot 6 7 for(int i=0; i<flowerbed.length; i++) 8 { 9 10 if(flowerbed[i] == 0) // if find empty spot, check its adjacent spots 11 { 12 if(i-1 >= 0 && flowerbed[i-1] == 1) // check left adjacent spot 13 continue; 14 if(i+1 < flowerbed.length && flowerbed[i+1] == 1) // check right adjacent spot 15 continue; 16 17 // if come here, meaning this 0 is valid spot to place flower 18 count++; 19 i++; // skip next 0 20 } 21 else if(flowerbed[i] == 1) // if find a flower spot, its next spot is not valid, skip it 22 { 23 i++; 24 } 25 26 27 if(count >= n) // if there are enough spots to place flower, no need to continue 28 return true; 29 30 } 31 32 return false; 33 } 34 }
参考资料：N/A
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