LeetCode 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.

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今天第一次做Leetcode的weekly contest，AC了三题，最后一题看起来好繁琐，没A。 第一题，简单贪心即可，从第0个位置开始，检查每个位置的前后是否为1，如果不为1，说明该位置可以种花，填上1，最后统计新填上的1的个数和n的关系。 代码如下： [cpp] class Solution { public: bool canPlaceFlowers(vector<int>& flowerbed, int n) { int m = flowerbed.size(); for (int i = 0; i < m; ++i) { if (flowerbed[i] == 0) { if (i – 1 >= 0 && flowerbed[i – 1] == 1)continue; if (i + 1 < m&&flowerbed[i + 1] == 1)continue; flowerbed[i] = 1; –n; } } return n <= 0; } }; [/cpp] 本代码提交AC，用时26MS。]]>