难度:Easy
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 = 1Output: TrueExample 2:Input: flowerbed = [1,0,0,0,1], n = 2Output: FalseNote: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.
可以创造一个以10开始的并以01结尾的数组,然后比较每两个1之间的间隔,可以放几个flower。比较其与n的大小。 例子中,假象start的索引为-2
class Solution {public:bool canPlaceFlowers(vector<int>& flowerbed, int n) {flowerbed.push_back(0);flowerbed.push_back(1);int total=0;int start=-2;for(int i=0; i<flowerbed.size();i++ ){if(flowerbed[i] ) {int r=i-start-3;start=i;// cout<<r<<endl;if(r<1 ) continue;total += (r+ r%2) /2;}}// cout<<total<<endl;return total>=n;}};
执行用时 :20 ms, 在所有 C++ 提交中击败了79.42%的用户 内存消耗 :11.2 MB, 在所有 C++ 提交中击败了17.02%的用户