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:

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.

    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;ii)
   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 };