The Solution to Leetcode 605 Can Place Flowers

Question:

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 nnew flowers can be planted in it without violating the no-adjacent-flowers rule.

假设你有一个长的花坛，其中一些地块种植，有些不是。 然而，花朵不能种植在相邻的地块上 - 他们会争取水，两者都会死亡。
给定一个花坛（表示为包含0和1的数组，其中0表示空，1表示不为空），数字n，如果可以种植n个新鲜花，则不返回无相邻花规则。

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.

注意：输入数组不会违反无相邻花规则。
           输入数组大小在[1，20000]的范围内。
           n是不会超过输入数组大小的非负整数。

思路：首先判断对于给定的数组，能够放多少数字1进去。有两个相邻的数字1就返回false.对于是数字0的元素，要检查左右两边的数字是不是0，如果是0，就可以放1，对于第一个数字，只用判断右边的数字是不是0，对于最后一个数字，只用判断左边的数字是不是0.最后求得一共能放进去1的数字的个数q与输入的n对比，如果q大于等于n，那么就返回true，否则false。

Answer:

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

Run code result:

Your input

[1,0,0,0,1]
2

Your answer

false

Expected answer

false