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:
- 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.
判断一个花床中可否放置数量为n盆花。要求花与花之间必须有一个空的间隔。
也就是说除了边界,必须有连续3个值都为0才可以放置1盆花。如[1,0,0,0,1]
这时需要判断边界值,因为在[0,0,1,0]或[0,1,0,0]这种情况下也可以放置1盆花。所以要在flowerbed的首位各放置一个0值来保证在这种情况下成立。
需要注意的是对flowerbed的遍历是从1~n-1。
class Solution { public: bool canPlaceFlowers(vector<int>& flowerbed, int n) { flowerbed.insert(flowerbed.begin(), 0); flowerbed.push_back(0); for (int i = 1; i < flowerbed.size() - 1; i++) { if (flowerbed[i - 1] + flowerbed[i] + flowerbed[i + 1] == 0) { i++; n--; } } return n <= 0; } }; // 16 ms