leetcode 605. 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.

我的方法是枚举

class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        int len = flowerbed.size();
        int num = 0;
        for (int i = 0; i < len; i +=2) {
            if (flowerbed[i] == 0) {
                if ( (i+1 >=len ||flowerbed[i + 1] == 0) && (i-1 < 0 ||flowerbed[i - 1] == 0) ) num++;
            }
        }
        if (num >= n) return true;
        num = 0;
        for (int i = 1; i < len; i += 2) {
            if (flowerbed[i] == 0) {
                if ((i+1 >= len || flowerbed[i + 1] ==0) && (i-1 < 0 || flowerbed[i - 1] == 0)) num++;
            }
        }
        if (num >= n) return true;
        return false;
    }
};

网上看了个不错的思路如下：

class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        int result = 0;
        int count = 1;
        // edge case: left and right
        for (int i = 0; i < flowerbed.size(); i++) {
            if (flowerbed[i] == 0) {
                count++;
            } else {
                result += (count - 1) / 2;
                count = 0;
            }
        }
        if (count != 0) {
            result += count / 2;
        }
        return result >= n;
    }
};