Leetcode Index

605.Can Place Flowers

605.Can Place Flowers

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 = 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.

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