LeetCode #605 CanPlaceFlowers. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
문제 좀 이상함,,
Weird problem,,
Runtime: 16 ms, faster than 93.88% of C++ online submissions for Can Place Flowers.
Memory Usage: 10.4 MB, less than 61.04% of C++ online submissions for Can Place Flowers.
LeetCode #605
Q.
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.
긴 꽃밭에 어느 부분은 모종이 심어져있고 어느 부분은 안심어져있다. 하지만, 바로 옆에는 모종을 심을 수 없다 - 물을 경쟁해서 둘다 죽게 된다.
주어진 꽃밭에서 ( 0 과 1 을 포함하는 배열로 나타나는데, 0은 비어있는 것이고 1은 비어있지 않은 것이다), 정수 n 과 함께 주어지는데, n 개의 꽃들이 죽지않고 심어질 수 있는지 확인하는 함수를 짜라.
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.
Process
// Process
//1. Input flowerbed and int n
//2. Iterate from begin to the end (flowerbed)
// 2.1. Check if it's 0 or 1
// 2.1.1. If it's 1 -> next 2
// 2.1.2. If it's 0
// 2.1.2.1. Check if it's adjacent 1
// 2.1.2.1.1. If not -> count
//3. Return count
// 처리과정
//1. 꽃밭배열과 정수를 입력받는다.
//2. 시작부터 끝까지 반복한다.
// 2.1. 0이나 1인지 확인해서
// 2.1.1. 1이면 -> 다다음으로 간다.
// 2.1.2. 0 이면
// 2.1.2.1. 주변에 1 이 있는지 확인해서
// 2.1.2.1.1. 없으면 카운트 센다.
//3. 결과 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
bool canPlaceFlowers(vector<int>& flowerbed, int n) {
bool result = false;
int count = 0;
for (int i = 0; i < flowerbed.size(); ++i) {
if (flowerbed[i] == 0) {
if ((i == 0 || flowerbed[i - 1] == 0) &&
(i == flowerbed.size()-1 || flowerbed[i + 1] == 0)) {
++count;
++i;
}
}
else {
++i;
}
}
if (count >= n)
result = true;
return result;
}
};
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