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.
Given an integer array flowerbed containing 0’s and 1’s, where 0 means empty and 1 means not empty, and an integer n, return if n new flowers can be planted in the flowerbed without violating the no-adjacent-flowers rule.

下面是普通的貪心做法，需要考慮三種邊界條件，時間復雜度還可以繼續優化.

優化前代碼

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

倘若當前位置是有花的或者是剛種下花，說明下一個位置不能種花，可以直接跳過。

優化后代碼

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

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