605. Can Place Flowers零一间隔种花

［抄题］：

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

 [暴力解法]：

时间分析：

空间分析：

 [优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

1. for循环一般还是从0到n, 如果怀疑，先做个标记，后续再改

［思维问题］：

for循环之内应该满足一个第一步的初始条件，再进行后续操作

［一句话思路］：

正常操作，直面灵魂拷问

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

1. 最多能种的花要比实际给的花的额度更大，count >= n

［二刷］：

［三刷］：

［四刷］：

［五刷］：

  [五分钟肉眼debug的结果]：

［总结］：

［复杂度］：Time complexity: O(n) Space complexity: O(1)

［英文数据结构或算法，为什么不用别的数据结构或算法］：

［关键模板化代码］：

正常操作，直面灵魂拷问

return count >= n;

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

 [代码风格] ：

class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
        //cc
        if (flowerbed == null || flowerbed.length == 0) {
            return false;
        }
        
        //ini
        int count = 0;
        
        //for loop as normal
        for (int i = 0; i < flowerbed.length && count <= n; i++) {
            if (flowerbed[i] == 0) {
                int prev = (i == 0) ? 0 : flowerbed[i - 1];
                int next = (i == flowerbed.length - 1) ? 0 : flowerbed[i + 1];
                
                if (prev == 0 && next == 0) {
                    count++;
                    flowerbed[i] = 1;
                }
            }
        }
        
        return count >= n;
    }
}