605. Can Place Flowers

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

含义：这道题给了我们一个01数组，其中1表示已经放了花，0表示可以放花的位置，但是有个限制条件是不能有相邻的花。为给定的n朵花能否放置完毕

解法1

 1     public boolean canPlaceFlowers(int[] flowerbed, int n) {
 2         int count = 0;
 3         for(int i = 0; i < flowerbed.length && count < n; i++) {
 4             if(flowerbed[i] == 0) {
 5          //get next and prev flower bed slot values. If i lies at the ends the next and prev are considered as 0. 
 6                int next = (i == flowerbed.length - 1) ? 0 : flowerbed[i + 1]; 
 7                int prev = (i == 0) ? 0 : flowerbed[i - 1];
 8                if(next == 0 && prev == 0) {
 9                    flowerbed[i] = 1;
10                    count++;
11                }
12             }
13         }
14         
15         return count == n;
16     }

解法2

 1        if (n == 0) return true;
 2         if (flowerbed.length == 0) return false;
 3         if (flowerbed.length == 1) {
 4             if (flowerbed[0] == 1) {
 5                 return false;
 6             }
 7             return n > 1 ? false : true;
 8         }
 9         if (flowerbed.length == 2) {
10             if (n == 1) {
11                 if (flowerbed[0] + flowerbed[1] == 0) return true;
12                 return false;
13             }
14             return false;
15         }
16         if (flowerbed[0] + flowerbed[1] == 0) {
17             flowerbed[0] = 1;
18             n--;
19         }
20         for (int i = 1; i <= flowerbed.length - 2; i++) {
21             if (flowerbed[i - 1] + flowerbed[i] + flowerbed[i + 1] == 0) {
22                 flowerbed[i] = 1;
23                 n--;
24             }
25         }
26 
27         if (n > 0) {
28             if (flowerbed[flowerbed.length - 2] + flowerbed[flowerbed.length - 1] == 0) {
29                 flowerbed[flowerbed.length - 1] = 1;
30                 n--;
31             }
32         }
33 
34         return n > 0 ? false : true;