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.
A:
-------------就是一个个对比---------
class Solution { public: bool canPlaceFlowers(vector<int>& flowerbed, int n) { bool isBeforeNoFlower = true; int planted = 0; for(int i=0;i< flowerbed.size(); ++i) { if(flowerbed[i] == 1){ isBeforeNoFlower = false; }else{ if(isBeforeNoFlower){ // before condition meet if(i == flowerbed.size()-1 || flowerbed[i+1]==0) // check next step condition { ++planted; isBeforeNoFlower = false; } }else{ isBeforeNoFlower = true; } } if(planted>=n)//canot check inside if statement, thus we need also check here return true; } return false; } };
Mistakes:
1: 忘了要考虑后面的位置
2: 没有考虑 n==0 的情况, 因此, 最后也需要考虑
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