Leetcode 605. Can Place Flowers

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本文深入解析LeetCode第605题CanPlaceFlowers的解题思路，探讨如何在遵循不相邻原则下种植花朵，通过贪心算法实现最优解，给出两种不同实现方式的代码示例。

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

题目大意：
给定一个只有0和1的数组，判断在1不能相邻的前提下是否能将数组内的n个0变成1.

解题思路：
采用贪心策略，只要出现连续3个0，就把中间的0变为1，需注意第一个元素(例如开头 [0，0，1 )或最后一个元素(例如 1，0，0])的特殊情况。时间复杂度为O(n)。代码1使用了额外空间O(2)，代码2不需要。

代码1：

class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        flowerbed.insert(flowerbed.begin(), 0);
        flowerbed.push_back(0);
        for(int i = 1; i < flowerbed.size() - 1; i++){
            if(flowerbed[i-1] != 1 && flowerbed[i+1] != 1 && flowerbed[i] != 1){
                flowerbed[i] = 1;
                n--;
            }
        }
        return n <= 0;
    }
};

代码2：

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