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

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.

思路：這是一道規律題

兩個1之間0的個數和可插入的1個數相關，相關關係為sum = (count-1)/2，count為兩個1之間區域性0的個數，sum為這一區域性可插入1的個數。

注意兩個邊界，如001可插入1,01則不可插入，解決辦法是兩側補零。

class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
        //為了防止001001,100100，10,01這幾種特殊情況，在數列兩端各加一個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;//記錄全域性可容納數
        int count = 0;//記錄區域性0的個數，區域性可容納數 = (區域性0的個數 - 1 )/2，
        //101 ：0
        //1001 ：0
        //10001 ：1
        //100001 ：1
        //1000001 ： 2
        //10000001 ： 2
        //...規律如上，零的個數：可容納數
        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;//最後一組區域性在上面for迴圈中未新增，新增上。如100100
        return (n<=sum)?true:false;
    }
}