LeetCode #852 PeakIndexInAMountainArray. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
STL 사용이 전체를 다 돌더라도 빠를 줄 알았는데, 그냥 for 문으로 절반만 확인하는게 더 빠름
I thought using STL will have better performance even though it has to iterate all element, but just checking half using for statement is better than STL.
Runtime: 12 ms, faster than 96.73% of C++ online submissions for Peak Index in a Mountain Array.
Memory Usage: 9.3 MB, less than 83.56% of C++ online submissions for Peak Index in a Mountain Array.
LeetCode #852
Q.
Let's call an array A a mountain if the following properties hold:
- A.length >= 3
- There exists some 0 < i < A.length - 1 such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]
Given an array that is definitely a mountain, return any i such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1].
배열 A 를 산이라고 가정하고, 배열 길이는 3보다 크거나 같고, 산의 피크(정상)을 기준으로 정상까지는 값이 쭉 올라가는 형태이고, 정상부터는 아래로 쭉 내려가는 형태이다. e.g. 1 2 3 4 5 4 3 2 ( index 4 peak 의 값 5 기준 )
Example 1:
Input: [0,1,0]
Output: 1
Example 2:
Input: [0,2,1,0]
Output: 1
Note:
- 3 <= A.length <= 10000
- 0 <= A[i] <= 10^6
- A is a mountain, as defined above.
Process
// Process
// 1. Input vector
// 2. Iterate from begin to the end
// 2.1. Check if it's bigger than maxVal
// 2.1.1. If so -> swap maxVal and store peakIndex
// 3. return peakIndex;
// 처리과정
//1. 배열 입력받는다.
//2. 시작부터 끝까지 반복한다.
// 2.1. 현재까지 나온 최대값보다 큰지 확인해서
// 2.1.1. 더 크면 -> 최대값 바꾸고, 피크인덱스로 저장해둔다.
//3. 피크인덱스 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
int peakIndexInMountainArray(vector<int>& A) {
//1
// return distance(A.begin(), max_element(A.begin(), A.end()));
//2
int peakIndex = 0;
int maxVal = -1;
bool isDone = false;
for (int i = 0; !isDone && i < A.size(); ++i) {
if (A[i] > maxVal)
{
maxVal = A[i];
peakIndex = i;
}
else
{
isDone = true;
}
}
return peakIndex;
}
};
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