leetcode 605) Can Place Flowers

LV. Easy 😎

https://leetcode.com/problems/can-place-flowers/

Can Place Flowers - LeetCode

 

문제

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.

 

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

 

Constraints:

• 1 <= flowerbed.length <= 2 * 10^4
• flowerbed[i] is 0 or 1.
• There are no two adjacent flowers in flowerbed.
• 0 <= n <= flowerbed.length

 

문제 해결법

앞에서부터 배열을 확인하면서 i - 1, i, i + 1이 0이면 i를 0으로 바꿔주고, n--를 해줬다.

배열이 끝나기 전에 n이 0이 되면 true를 리턴!

 

배열의 사이즈가 1이거나 n이 0인 경우는 특별 케이스로 위에 빼줘서 계산해줬다.

 

해결 코드

class Solution {
public:
	bool canPlaceFlowers(vector<int>& flowerbed, int n) {
		int length = flowerbed.size();

		if (n == 0 || (length == 1 && flowerbed[0] == 0 && n <= 1))
			return 1;
		else if (length == 1)
			return 0;
		for (int i = 0; i < length; ++i) {
			cout << i << endl;
			if (i == 0 && flowerbed[i] == 0 && flowerbed[i + 1] == 0) {
				flowerbed[i] = 1;
				i++;
				n--;
				if (n == 0)
					return 1;
			}
			else if (i == 0)
				continue;
			else if (i == length - 1 && flowerbed[i - 1] == 0 && flowerbed[i] == 0) {
				flowerbed[i] = 1;
				i++;
				n--;
				if (n == 0)
					return 1;
			}
			else if (flowerbed[i - 1] == 0 && flowerbed[i] == 0 && flowerbed[i + 1] == 0){
				flowerbed[i] = 1;
				i++;
				n--;
				if (n == 0)
					return 1;
			}
		}
		return n == 0;
	}
};