[leetcode]: 605. Can Place Flowers

1.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.
用一维数组表示要种花的花床，0表示未种，1表示已种。相邻两处不能同时种花。给出花床的数组和需要种花的数量n，判断是否能把n个花种下去。

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

2.分析

相邻位置不能同时种花，那么可以种花分三种情况：
1）数组首部 00x 头部的0可以种花
2）数组中间 000 中间的0可以种花
3）数组尾部 x00 尾部的0可以种花
遍历一遍数组即可。

3.代码

class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        int length = flowerbed.size();
        int i = 0;
        for (; i < length&&n > 0; i++) {
            if (flowerbed[i] == 0) {
                int left = i - 1 > 0 ? i - 1 : 0;
                int right = i + 1 < length - 1 ? i + 1 : length - 1;
                if (flowerbed[left] == 0 && flowerbed[right] == 0) {
   //左右两边都没有种花
                    flowerbed[i] = 1;
                    --n;
                }
            }
        }
        return n == 0;
    }
};