C#LeetCode刷题之#605-种花问题( Can Place Flowers)

C#LeetCode刷题之#605-种花问题( Can Place Flowers)

问题

假设你有一个很长的花坛,一部分地块种植了花,另一部分却没有。可是,花卉不能种植在相邻的地块上,它们会争夺水源,两者都会死去。

给定一个花坛(表示为一个数组包含0和1,其中0表示没种植花,1表示种植了花),和一个数 n 。能否在不打破种植规则的情况下种入 n 朵花?能则返回True,不能则返回False。

输入: flowerbed = [1,0,0,0,1], n = 1

输出: True

输入: flowerbed = [1,0,0,0,1], n = 2

输出: False

注意:

数组内已种好的花不会违反种植规则。
输入的数组长度范围为 [1, 20000]。
n 是非负整数,且不会超过输入数组的大小。

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.

Input: flowerbed = [1,0,0,0,1], n = 1

Output: True

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.

示例

public class Program {

    public static void Main(string[] args) {
        int[] nums = null;

        nums = new int[] { 1, 0, 0, 0, 1 };
        var res = CanPlaceFlowers(nums, 1);
        Console.WriteLine(res);

        Console.ReadKey();
    }

    private static bool CanPlaceFlowers(int[] flowerbed, int n) {
        //该题比较简单,处理好边界即可
        //需要种植的花为0,总是可以
        if(n == 0) return true;
        //数组为0,不可种植
        if(flowerbed.Length == 0) {
            return false;
        }
        //当数组为1时,根据是否种植直接判定即可
        if(flowerbed.Length == 1) {
            if(flowerbed[0] == 0) {
                return true;
            } else {
                return false;
            }
        }
        //记录可以种植的数量
        int count = 0;
        //前2个分析
        if(flowerbed.Length >= 2 &&
           flowerbed[0] == 0 &&
           flowerbed[1] == 0) {
            flowerbed[0] = 1;
            count++;
        }
        //连续3个为0时,可种植
        for(int i = 1; i < flowerbed.Length - 1; i++) {
            if(flowerbed[i - 1] == 0 &&
               flowerbed[i] == 0 &&
               flowerbed[i + 1] == 0) {
                count++;
                flowerbed[i] = 1;
            }
        }
        //后2个分析
        if(flowerbed.Length >= 2 &&
           flowerbed[flowerbed.Length - 1] == 0 &&
           flowerbed[flowerbed.Length - 2] == 0) {
            flowerbed[flowerbed.Length - 1] = 1;
            count++;
        }
        //可种植数大于等于即将种植的数量时,返回true
        if(count >= n) return true;
        //无解时,返回false
        return false;
    }

}

以上给出1种算法实现,以下是这个案例的输出结果:

True

分析:

该题比较简单,若使用其它ADT,可以考虑左右边界补0的方法,这样可以少处理部分边界问题。

显而易见,以上算法的时间复杂度为: O(n)

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