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:
我的方法是枚举
class Solution { public: bool canPlaceFlowers(vector<int>& flowerbed, int n) { int len = flowerbed.size(); int num = 0; for (int i = 0; i < len; i +=2) { if (flowerbed[i] == 0) { if ( (i+1 >=len ||flowerbed[i + 1] == 0) && (i-1 < 0 ||flowerbed[i - 1] == 0) ) num++; } } if (num >= n) return true; num = 0; for (int i = 1; i < len; i += 2) { if (flowerbed[i] == 0) { if ((i+1 >= len || flowerbed[i + 1] ==0) && (i-1 < 0 || flowerbed[i - 1] == 0)) num++; } } if (num >= n) return true; return false; } };
网上看了个不错的思路如下：
class Solution { public: bool canPlaceFlowers(vector<int>& flowerbed, int n) { int result = 0; int count = 1; // edge case: left and right for (int i = 0; i < flowerbed.size(); i++) { if (flowerbed[i] == 0) { count++; } else { result += (count - 1) / 2; count = 0; } } if (count != 0) { result += count / 2; } return result >= n; } };
leetcode 605. Can Place Flowers
原文：http://www.cnblogs.com/pk28/p/7253881.html