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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0101_0200.s0119_pascals_triangle_ii;

// #Easy #Array #Dynamic_Programming #Data_Structure_II_Day_3_Array #Dynamic_Programming_I_Day_12
// #Udemy_Dynamic_Programming #2022_06_23_Time_0_ms_(100.00%)_Space_41.5_MB_(70.65%)

import java.util.ArrayList;
import java.util.List;

/**
 * 119 - Pascal's Triangle II\.
 *
 * Easy
 *
 * 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:
 *
 * ![](https://upload.wikimedia.org/wikipedia/commons/0/0d/PascalTriangleAnimated2.gif)
 *
 * **Example 1:**
 *
 * **Input:** rowIndex = 3
 *
 * **Output:** [1,3,3,1] 
 *
 * **Example 2:**
 *
 * **Input:** rowIndex = 0
 *
 * **Output:** [1] 
 *
 * **Example 3:**
 *
 * **Input:** rowIndex = 1
 *
 * **Output:** [1,1] 
 *
 * **Constraints:**
 *
 * *   `0 <= rowIndex <= 33`
 *
 * **Follow up:** Could you optimize your algorithm to use only `O(rowIndex)` extra space?
**/
public class Solution {
    public List getRow(int rowIndex) {
        int[] buffer = new int[rowIndex + 1];
        buffer[0] = 1;
        computeRow(buffer, 1);
        // Copy buffer to List of Integer.
        List ans = new ArrayList<>(buffer.length);
        for (int j : buffer) {
            ans.add(j);
        }
        return ans;
    }

    private void computeRow(int[] buffer, int k) {
        if (k >= buffer.length) {
            return;
        }
        int previous = buffer[0];
        for (int i = 1; i < k; i++) {
            int tmp = previous + buffer[i];
            previous = buffer[i];
            buffer[i] = tmp;
        }
        buffer[k] = 1;
        computeRow(buffer, k + 1);
    }
}