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

 

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.

种花=-=，不能相邻种，不然会挂掉，我的思路就是考虑各种情况。。。

 1 class Solution {
 2 public:
 3     bool canPlaceFlowers(vector<int>& flowerbed, int n) {
 4         if (find(flowerbed.begin(), flowerbed.end(), 1) == flowerbed.end())
 5         {//if there was no flower
 6             return (flowerbed.size() + 1) / 2 >= n;
 7         }
 8 
 9         int size = flowerbed.size();
10         int start = 0, end = 0, count = 0;
11         int max = 0;
12         for (int i = 0; i < size; ++i)
13         {
14             if (flowerbed[i] == 0)
15             {
16                 ++count;
17             }
18             else
19             {
20                 max += count / 2;
21                 start = count;
22                 count = 0;
23                 break;
24             }
25         }
26         
27         for (int i = size - 1; i >= 0; --i)
28         {
29             if (flowerbed[i] == 0)
30             {
31                 ++count;
32             }
33             else
34             {
35                 max += count / 2;
36                 end = size - count;
37                 count = 0;
38                 break;
39             }
40         }
41         
42         for (int i = start; i < end; ++i)
43         {
44             if (flowerbed[i] == 0)
45             {
46                 ++count;
47             }
48             else
49             {
50                 max += (count - 1) / 2;
51                 count = 0;
52             }
53         }
54 
55         return max >= n ? true : false;
56     }
57 };

贴一个技巧性强一点的discuss的解答，很给力=。=

 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 };

 

转载于:https://www.cnblogs.com/jiadyang/p/8654278.html