605. Can Place Flowers

題目連結

605.1 題目描述：

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.

605.2 解題思路：

1. 思路一：兩步一處理，開頭和結尾分開處理。遍歷flowerbed，首先，如果i==0,並且f[i]==0，f[i+1]==0，則n–，標記0個數的flag=0。然後進入迴圈，如果f[i]==1，則flag=0；否則，flag++。如果flag==2，且f[i+1]==0，則n–，flag=0。最後i++。遍歷結束，如果到結尾，flag==1，且f[i]==0，n–。然後返回n<+0。即可。思想就是兩個一組同為0來討論。

2. 思路二：三個一組來討論。正常思路都應該這樣！遍歷flowerbed，如果f[i]==0，i==0或者f[i-1]==0，i==f.size-1或者f[i+1]==0，即這三個條件同時滿足，說明可以種花，則f[i]==1，為了下一個條件判斷準備，然後n–。當遍歷結束後，如果n<=0，即花種完了，則返回true，否則false。

605.3 C++程式碼：

1、思路一程式碼（25ms）：

class Solution120 {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        if (n <= 0)
            return false;
        if (flowerbed.size() == 1)
        {
            if (flowerbed[0] == 0 && n == 1)
                return true;
            else
                return false;
        }
        if (flowerbed.size() == 2)
        {
            if (flowerbed[0] == 0 && flowerbed[1] == 0 && n == 1)
                return true;
            else
                return false;
        }
        if (n > (flowerbed.size() / 2 + 1))
            return false;
        int flag1 = 0;
        int i = 0;
        if (flowerbed[i] == 0 && flowerbed[i+1] == 0)
        {
            n--;
            flag1 = 0;
            i++;
        }
        while(i < flowerbed.size()-1)
        {
            if (flowerbed[i]==1)
                flag1 = 0;
            else
                flag1++;        
            if (flag1 == 2 && flowerbed[i+1] == 0)
            {
                n--;
                flag1 = 0;
            }
            i++;
        }
        if (flag1 == 1 && flowerbed[i] == 0)
            n--;
        return n<=0;
    }
};

2、思路二程式碼（19ms）

class Solution120_1 {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        for (int i = 0; i < flowerbed.size();i++)
        {
            if (flowerbed[i] == 0 && (i == 0 || flowerbed[i - 1] == 0) && (i == flowerbed.size() - 1 || flowerbed[i + 1] == 0))
            {
                flowerbed[i] = 1;
                n--;
            }
        }
        return n <= 0;
    }
};

605.4 Python程式碼：

1、思路二程式碼（59ms）

class Solution(object):
    def canPlaceFlowers(self, flowerbed, n):
        """
        :type flowerbed: List[int]
        :type n: int
        :rtype: bool
        """
        for i in range(len(flowerbed)):
            if flowerbed[i]==0 and (i==0 or flowerbed[i-1]==0) and (i==len(flowerbed)-1 or flowerbed[i+1]==0):
                flowerbed[i]=1
                n-=1
        return n<=0