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 nnew 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.

 1 class Solution {
 2 public:
 3     bool canPlaceFlowers(vector<int>& flowerbed, int n) {
 4         int tmp_n = 0;
 5         int num_zero = 0;
 6         bool left_no_one = true;
 7         bool right_no_one = false;
 8         int count = flowerbed.size();
 9         for(int i=0;i<count;++i)
10         {
11             if(flowerbed[i] == 0)//记录0的个数
12             {
13                 num_zero++;
14                 if(i == (count-1))//判断是不是最右侧没有1
15                 {
16                    right_no_one = true; 
17                 }
18             }
19             else//遇到1之后开始算之前的零够能种多少花
20             {
21                 if(left_no_one == true)//先判断是最左侧没有1
22                 {
23                     if(num_zero >= 2)
24                     {
25                         if(num_zero%2 == 0)
26                         {
27                  
28            num_zero += 1;
29                         }
30                         tmp_n += (((num_zero-3)/2)+1);
31                     }
32                     
33                 }
34                 else //左侧有1
35                 {
36                     if(num_zero >= 3)
37                     {
38                         if(num_zero%2 == 1)
39                         {
40                             num_zero +=1;
41                         }
42                         tmp_n += (((num_zero-4)/2)+1);
43                     }
44                 
45                     
46                 }
47                 num_zero = 0; 
48                 left_no_one = false;//清除左侧没有1的标志
49             }
50         }
51         if(left_no_one == true && right_no_one == true)
52         {
53             if(num_zero >= 1)
54             {
55                 if(num_zero%2 == 1)
57                 {
58                     num_zero += 1;
59                 }
60                 tmp_n += (((num_zero-2)/2)+1);
61             }
62             
63         }
64         else
65         {
66             if(num_zero >= 2)
67             {
68                 if(num_zero%2 == 0)
69                 {
70                     num_zero += 1;
71                 }
72                 tmp_n += (((num_zero-3)/2)+1);
73             }
74         }
75 
76         if(tmp_n >= n)
77         {
78             return true;
79 
80         }
81         else
82         {
83             return false;
84         }
85 
86     }
87 };