LeetCode #977 SquaresOfASortedArray. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
Runtime: 124 ms, faster than 23.05% of C++ online submissions for Squares of a Sorted Array.
Memory Usage: 13.4 MB, less than 68.04% of C++ online submissions for Squares of a Sorted Array.
It could be better performance if sort using absolute and square at the same time without stl ?
stl 안쓰고 직접 절대값비교로 정렬하면서 제곱해주면 퍼포먼스 잘 나올듯?
LeetCode #977
Q.
Given an array of integers A sorted in non-decreasing order, return an array of the squares of each number, also in sorted non-decreasing order.
주어진 정수배열 A는 오름차순으로 정렬되어있고, 각 수의 제곱한 배열을 리턴하는데, 오름차순으로 정렬하여 내보내라.
Example 1:
Input: [-4,-1,0,3,10]
Output: [0,1,9,16,100]
Example 2:
Input: [-7,-3,2,3,11]
Output: [4,9,9,49,121]
Note:
- 1 <= A.length <= 10000
- -10000 <= A[i] <= 10000
- A is sorted in non-decreasing order.
// Process
//1. Input vector
//2. Square all element
//3. Sort
//4. Return vector
// 처리과정
//1. 배열 입력받는다.
//2. 제곱한다.
//3. 정렬한다.
//4. 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
vector<int> sortedSquares(vector<int>& A) {
for (int i = 0; i < A.size(); ++i) {
A[i] = pow(A[i],2);
}
sort(A.begin(), A.end());
return A;
}
};
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