leetcode:(605) Can Place Flowers(java)

最新推荐文章于 2019-03-10 20:18:00 发布

Sunshine_liang1

最新推荐文章于 2019-03-10 20:18:00 发布

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本文链接：https://blog.youkuaiyun.com/Sunshine_liang1/article/details/83987046

本文探讨了在遵循不相邻种植规则的情况下，如何在由0和1组成的数组中找到是否可以种植n朵花的方法。通过遍历数组并检查每个位置及其邻居的状态，实现了对给定数量花卉的有效种植判断。

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题目：

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:

题目描述：

相邻的区间不能种植花卉，求是否可以种植下n朵花。

具体解题思路及代码如下：

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.

package Leetcode_Github;

public class GreetyThought_CanPlaceFlowers_605_1112 {
    public boolean CanPlaceFlowers(int[] flowerbed, int n) {
        //判断是否合法输入
        if (flowerbed == null || flowerbed.length == 0 || n > flowerbed.length) {
            return false;
        }

        //记录可以种植花朵的数量
        int count = 0;
        for (int i = 0; i < flowerbed.length && count < n; i++) {
             if(flowerbed[i] == 1){
                 continue;
             }
            //当前索引前面的数组值
            int pre = i == 0 ? flowerbed[i] : flowerbed[i - 1];
            //当前索引后面的数组值
            int next = i == flowerbed.length - 1 ? 0 : flowerbed[i + 1];

            //当前索引值不为1，并且当前索引的前、后数组值均为0，则可以种植花朵
            if (pre == 0 && next == 0 ) {
                count++;
                flowerbed[i] = 1;
            }

        }
        return count >= n;
    }
}