LeetCode #119 Pascal'sTriangle2. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접LeetCode #119 Pascal'sTriangle2. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접, 파스칼의 삼각형, Pascal, 파스칼
파스칼의삼각형1번 문제와 거의 유사하고, 퍼포먼스 100%, 메모리 50%
It's almost similar with Pascal's Triangle #1, Performance 100%, memory 50%
LeetCode #119
Q.
Given a non-negative index k where k ≤ 33, return the kth index row of the Pascal's triangle.
Note that the row index starts from 0.
양의정수 index k 는 33보다 작거나 같고, k번째 인덱스의 파스칼 삼각형을 반환해라.
로우 인덱스는 0부터 시작한다.
In Pascal's triangle, each number is the sum of the two numbers directly above it.
파스칼의 삼각형에서, 각 숫자는 바로 위의 두개 숫자 합이다.
e.g.
Example:
Input : 3
Output : [1,3,3,1]
Follow up:
Could you optimize your algorithm to use only O(k) extra space?
공간복잡도 O(k) 로 최적화 가능한가?
Process
// Process
//1. Input rowIndex
//2. Iterate from 0 to rowIndex
// 2.1. Iterate from 0 to length of that floor
// 2.1.1. Fill the tempVector in 1
// 2.1.2. Fill the tempVector in sum of 2 number in previous vector
// 2.1.3. If it's last one, then fill the tempVector in 1
// 2.2. Replace temp to previous vector
//3. Return result (previous)
// 처리과정
//1. rowIndex 입력받는다.
//2. 0부터 rowIndex 까지 반복한다.
// 2.1. 0부터 해당 층의 길이까지 반복한다.
// 2.1.1. tempVector 에 1을 채운다.
// 2.1.2. tempVector 에 이전 vector 에서 바로 위 2개 더한 값을 채운다.
// 2.1.3. 마지막이면, tempVector 에 1을 채운다.
// 2.2. tempVector 를 previous vector 로 치환한다.
//3. 결과 반환한다. (previousVector)
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
vector<int> getRow(int rowIndex) {
vector<int> resultVector;
if (rowIndex == 0) {
resultVector.push_back(1);
return resultVector;
}
for (int i = 0; i <= rowIndex; ++i) {
vector<int> tempVector;
for (int j = 0; j < i; ++j) {
if (j == 0) {
tempVector.push_back(1);
}
else {
tempVector.push_back(resultVector.at(j - 1) + resultVector.at(j));
}
if (j == i - 1) {
tempVector.push_back(1);
}
}
resultVector = tempVector;
}
return resultVector;
}
};
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