605. Can Place Flowers

luozhiyun

发布于 2019-09-10 18:45:53

2780

发布于 2019-09-10 18:45:53

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:

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.

我的答案:

class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < flowerbed.length; i++) {
            if(flowerbed[i] == 1){
                list.add(i);
            }
        }
        if(list.size()==0 ){
            n = n-(flowerbed.length+1)/2;
            return n<=0;
        }
        //head
        Integer head = list.get(0);
        if(head>1){
            n = n-head/2;
        }
        //tail
        Integer tail = list.get(list.size()-1);
        if(flowerbed.length-tail-1>1){
            n = n-(flowerbed.length-tail-1)/2;
        }
        
        for(int i = 0;i<list.size()-1;i++){
            int begin = list.get(i);
            int end = list.get(i+1);
            if(end-begin>2){
                n = n-(end-begin-2)/2;
            }
        }
        
        return n <= 0;
    }
}

大牛的答案:

public boolean canPlaceFlowers(int[] flowerbed, int n) {
    int count = 1;
    int result = 0;
    for(int i=0; i<flowerbed.length; i++) {
        if(flowerbed[i] == 0) {
            count++;
        }else {
            result += (count-1)/2;
            count = 0;
        }
    }
    if(count != 0) result += count/2;
    return result>=n;
}

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原始发表：2018-01-24 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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