[Leetcode] 605. Can Place Flowers 解题报告

最新推荐文章于 2019-09-13 15:51:52 发布

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解决一个经典的编程问题，即在遵循不相邻原则的情况下，在给定的花坛中判断是否可以种植指定数量的新花。该问题通过遍历数组并检查每个空位是否符合种植条件来解决。

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题目：

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.

思路：

热身题目，哈哈。

代码：

class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        int m = 0;
        for (int i = 0; i < flowerbed.size(); ++i) {
            bool empty = flowerbed[i] == 0;
            bool left_empty = (i == 0) || flowerbed[i - 1] == 0;
            bool right_empty = (i + 1 == flowerbed.size()) || flowerbed[i + 1] == 0;
            if (empty && left_empty && right_empty) {
                flowerbed[i] = 1;
                ++m;
            }
        }
        return m >= n;
    }
};