leetcode -- 605. Can Place Flowers 【边界处理 + 数学规律】

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:

1. The input array won't violate no-adjacent-flowers rule.
2. The input array size is in the range of [1, 20000].
3. n is a non-negative integer which won't exceed the input array size.

题意

种花，两个花不能相邻（因竞争而死），给定一定的花床，判断其是否仍然可种下给定数量的花。

解法1：（暴力方法）

• 【判断可种的规则】对于介于 0 和（ 数组长度-1）之间的元素，判定规则需考虑：元素自身，元素的左右。
• 【边界规则】针对0，没有左元素。
• 【边界规则】针对（ 数组长度-1），没有右有元素。
  

public class Solution {
	public boolean canPlaceFlowers(int[] flowerbed, int n) {
		
		if(flowerbed.length == 1){
			if(flowerbed[0] == 0){
				n--;
			}
			return n <= 0;
		}
		
		for(int i = 0;i < flowerbed.length;i++){
			if(i == 0){
				if(!(flowerbed[i] ==1) && !(flowerbed[i+1] == 1)){
					flowerbed[i] = 1;
					n--;
				}
				continue;
			}
			
			if(i == flowerbed.length -1){
				if(!(flowerbed[i] ==1) && !(flowerbed[i-1] == 1)){
					flowerbed[i] = 1;
					n--;
				}
				continue;
			}
			
			if(!(flowerbed[i] ==1) && !(flowerbed[i-1] == 1) && !(flowerbed[i+1] == 1)){
				flowerbed[i] = 1;
				n--;
			}
			
		}
		return n <= 0;
	}
}

解法2：（利用数学规律）

• 【统计连续空地】使用count来进行统计。
• 【统计连续空地上可种】(count -1) / 2
• 【特殊处理 + 最后一个count】 count / 2
  

public boolean canPlaceFlowers(int[] flowerbed, int n) {
    int count = 1;
    int result = 0;
    for(int i=0; i=n;
}