Can Place Flowers

投稿日 |

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.

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Example 1:
Input: flowerbed = [1,0,0,0,1], n = 1
Output: True
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Example 2:
Input: flowerbed = [1,0,0,0,1], n = 2
Output: False

Note:

  • The input array won’t violate no-adjacent-flowers rule.
  • The input array size is in the range of [1, 20000].
  • n is a non-negative integer which won’t exceed the input array size.

Method

从头到尾遍历一遍，如果前面的节点和后面的节点的值都为0，则这个点可以种花。然后将这个节点的值设为1。

Solution

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public class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
        int count = 0;
        for(int i = 0; i < flowerbed.length; i++){
            int prev = (i - 1) > -1 ? flowerbed[i-1] : 0;
            int next = (i + 1) < flowerbed.length ? flowerbed[i+1] : 0;
            if(prev == 0 && next == 0 && flowerbed[i] != 1){
                flowerbed[i] = 1;
                count++;
            }
        }
        return count >= n;
    }
}