(LeetCode) Can Place Flowers - LeetCode

Description:

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.

 

Solution:

class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
        
        if(flowerbed==null||flowerbed.length==0){
            return false;
        }
        
        int k = 0;
        for(int i = 0; i<flowerbed.length; i++){
            
            if(checkAdjacent(flowerbed,i)){
                k=k+1;
                flowerbed[i] =1;
            }
        }
        
        if(k>=n){
            return true;
        }
        else{
            return false;
        }
    }
    
    public static boolean checkAdjacent(int[] flowerbed, int i){
        
            if(flowerbed.length==1&& flowerbed[i]==0){
                return true;
            }
        
            if(flowerbed[i]==1){
                return false; 
            }
            else{
                if (i ==0){
                
                if(i+1<flowerbed.length && flowerbed[i+1]!=1){
                    return true;
                }
             }
        
            else if(i ==flowerbed.length-1){
                
                if(i-1>=0&&flowerbed[i-1]!=1){
                    return true;
                }
            }
        
            else{
                if(i+1<flowerbed.length&i-1>=0&&flowerbed[i+1]!=1&&flowerbed[i-1]!=1){
                    return true;
                }
                
            }
                return false;
                
          }
            
        
    }
}

 

posted @ 2019-09-03 17:57  CodingYM  阅读(140)  评论(0编辑  收藏  举报