[LeetCode 119] Pascal's Triangle II

문제 바로가기

Description

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Given an integer rowIndex, return the rowIndexth (0-indexed) row of the Pascal’s triangle.

In Pascal’s triangle, each number is the sum of the two numbers directly above it as shown:

Example 1

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• Input: rowIndex = 3
• Output: [1,3,3,1]

Example 2

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• Input: rowIndex = 0
• Output: [1]

Example 3

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• Input: rowIndex = 1
• Output: [1,1]

Constraints

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• 0 <= rowIndex <= 33

Code

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내 제출

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class Solution {
    public List<Integer> getRow(int rowIndex) {
        int[][] triangle = new int[100][100]; // 실제 연산용 변수
        List<Integer> result = new ArrayList<>(); // 리턴용 변수

        for (int i = 0; i < (rowIndex + 1); i++) {
            triangle[i][0] = 1; // 초기값 지정
            triangle[i][i] = 1; // ,,
            for (int j = 1; j < i; j++) {
                triangle[i][j] = triangle[i - 1][j - 1] + triangle[i - 1][j];
            }
        }

        for (int i = 0; i < triangle.length; i++) { // rowIndex의 값 저장
            if (i == rowIndex) {
                for (int j = 0; j < (i + 1); j++) {
                    result.add(triangle[i][j]);
                }
            }
        }

        return result;
    }
}

Runtime	Memory
1 ms	42.5 MB

다른 풀이

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class Solution {
    public long findNcR(int n , int r){
        long res = 1;
        for(int i = 0; i < r; i++){
            res = res*(n-i);
            res = res/(i+1);
        }
        return res;
    }
    public List<Integer> getRow(int rowIndex) {
        List<Integer> ans = new ArrayList<>();
        for(int i = 0; i <= rowIndex ; i++){
            ans.add((int)findNcR(rowIndex , i));
        }
        return ans;
    }
}

Reference

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