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 };