605. Can Place Flowers【easy】

605. Can Place Flowers【easy】

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:

1. The input array won‘t violate no-adjacent-flowers rule.
2. The input array size is in the range of [1, 20000].
3. n is a non-negative integer which won‘t exceed the input array size.

错误解法：

 1 class Solution {
 2 public:
 3     bool canPlaceFlowers(vector<int>& flowerbed, int n) {
 4         int max = 0;
 5         int temp = 0;
 6
 7         for (int i = 0; i < flowerbed.size(); ++i)
 8         {
 9             if (flowerbed[i] == 0)
10             {
11                 temp++;
12                 max = temp > max ? temp : max;
13             }
14             else
15             {
16                 temp = 0;
17             }
18         }
19
20
21         if ((max - 2) % 2)
22         {
23             return ((max - 2) / 2 + 1 >= n);
24         }
25         else
26         {
27             return ((max - 2) / 2 >= n);
28         }
29     }
30 };

想要用最长连续的0去求，调试了很久，但还是条件判断有问题，这种方法不妥！

参考大神们的解法，如下：

解法一：

 1 public class Solution {
 2     public boolean canPlaceFlowers(int[] flowerbed, int n) {
 3         int count = 0;
 4         for(int i = 0; i < flowerbed.length && count < n; i++) {
 5             if(flowerbed[i] == 0) {
 6          //get next and prev flower bed slot values. If i lies at the ends the next and prev are considered as 0.
 7                int next = (i == flowerbed.length - 1) ? 0 : flowerbed[i + 1];
 8                int prev = (i == 0) ? 0 : flowerbed[i - 1];
 9                if(next == 0 && prev == 0) {
10                    flowerbed[i] = 1;
11                    count++;
12                }
13             }
14         }
15
16         return count == n;
17     }
18 }

解法二：

1 public boolean canPlaceFlowers(int[] flowerbed, int n) {
2     for (int idx = 0; idx < flowerbed.length && n > 0; idx ++)
3         if (flowerbed [idx] == 0 && (idx == 0 || flowerbed [idx - 1] == 0) && (idx == flowerbed.length - 1 || flowerbed [idx + 1] == 0)) {
4             n--;
5             flowerbed [idx] = 1;
6         }
7     return n == 0;
8 }

解法一和解法二都需要随时更新原来数组的状态

解法三：

 1 class Solution {
 2 public:
 3     bool canPlaceFlowers(vector<int>& flowerbed, int n) {
 4         flowerbed.insert(flowerbed.begin(),0);
 5         flowerbed.push_back(0);
 6         for(int i = 1; i < flowerbed.size()-1; ++i)
 7         {
 8             if(flowerbed[i-1] + flowerbed[i] + flowerbed[i+1] == 0)
 9             {
10                 --n;
11                 ++i;
12             }
13
14         }
15         return n <=0;
16     }
17 };

不更新原来数组的值，通过多移下标来实现