[leetcode] Can Place Flowers

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.

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分析：题目意思还是很清晰的，给定数组a，由0、1组成，规则是：1不能连续出现。要求在a中查找满足条件可以将多少个0变成1。

这里就需要考虑三种情况：开头、末尾和中间。

代码如下：

class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
        int count = 0;
        int i = 0;
        if ( flowerbed.length == 1){
            if ( flowerbed[0] == 1 ) {
                if ( n == 0 ) return true;
                else return false;
            }
            else 
                return n<=1;
        }
        while ( i < flowerbed.length ){
            if ( flowerbed[i] == 0 && (i==0||flowerbed[i-1]==0) && (i==flowerbed.length-1||flowerbed[i+1]==0) ){
                flowerbed[i]=1;
                count ++;
                i+=2;
                if ( i >= flowerbed.length ) break;
            }
            else{
                i ++;
            }
        }
        return count >= n;
        
    }
}

    运行时间6ms。

这个题目没什么太需要总结的，就是考虑好特殊情况，整理好思路就好了。