605. Can Place Flowers

题目来源【Leetcode】

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.

这道题就是任何任何两个1不能挨着，然后问你能否放所给的花的朵数，刚开始我的代码是这样的，运行时间比较长，因为要判断是否在两端的情况，如下：

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

后来，在头尾都插入一个0，就可以省事很多，运行时间也减少了许多，如下：

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