Can Place Flowers：数组相邻元素不为0的插入问题

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 = 1Output: True

Example 2:

Input: flowerbed = [1,0,0,0,1], n = 2Output: 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.

思路：这是一道规律题

两个1之间0的个数和可插入的1个数相关，相关关系为sum = (count-1)/2，count为两个1之间局部0的个数，sum为这一局部可插入1的个数。

注意两个边界，如001可插入1,01则不可插入，解决办法是两侧补零。

class Solution {    public boolean canPlaceFlowers(int[] flowerbed, int n) {        //为了防止001001,100100，10,01这几种特殊情况，在数列两端各加一个0        StringBuilder sb = new StringBuilder();        sb.append(0);        for(int i = 0;i<flowerbed.length;i++){            sb.append(flowerbed[i]);        }        sb.append(0);        String s = sb.toString();        int sum = 0;//记录全局可容纳数        int count = 0;//记录局部0的个数，局部可容纳数 = (局部0的个数 - 1 )/2，        //101 ：0        //1001 ：0        //10001 ：1        //100001 ：1        //1000001 ： 2        //10000001 ： 2        //...规律如上，零的个数：可容纳数        for(int i = 0 ;i < s.length(); i++){            if(s.charAt(i)=='1'){                sum += (count-1)/2;                count = 0;            }else{                count ++;            }        }        sum += (count -1) /2;//最后一组局部在上面for循环中未添加，添加上。如100100        return (n<=sum)?true:false;    }}