标签:represent ber erb color 而且 exce 计数器 tput ann
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:
算法:贪婪法。一个count记录当前种了多少了。for循环遍历一次,看到0的,而且前面是0或者头,而且后面是0或者尾,就可以种下去并加计数器。
细节:小心边界条件,比如就种0个花的。跑一遍改结构。
实现:
class Solution { public boolean canPlaceFlowers(int[] flowerbed, int n) { if (flowerbed == null || flowerbed.length == 0) { return false; } int count = 0; for (int i = 0; i < flowerbed.length; i++) { if (flowerbed[i] == 0 && (i == 0 || flowerbed[i - 1] == 0) && (i == flowerbed.length - 1 || flowerbed[i + 1] == 0)) { count ++; // 改变传入的数据,需要问是否允许 flowerbed[i] = 1; } if (count >= n) { return true; } } return false; } }
leetcode605- Can Place Flowers- easy
标签:represent ber erb color 而且 exce 计数器 tput ann
原文地址:http://www.cnblogs.com/jasminemzy/p/7920611.html