Leetcode Pascal's Triangle II problem solution

In this Leetcode Pascal's Triangle II problem solution we have Given an integer rowIndex, return the rowIndexth (0-indexed) row of Pascal's triangle. In Pascal's triangle, each number is the sum of the two numbers directly above it.

Problem solution in Python.

class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        if rowIndex == 0: 
            return [1]
        if rowIndex == 1: 
            return [1, 1]
        if rowIndex == 2: 
            return [1, 2, 1]
        
        ans = [1] * (rowIndex + 1)
        prev = [1, 2, 1]
        
        for i in range(3, rowIndex + 1):
            for j in range(1, i):
                ans[j] = prev[j] + prev[j - 1]
            prev = ans[:]
            
        return ans

Problem solution in Java.

public List<Integer> getRow(int rowIndex) {      
        List<Integer> result = new ArrayList<>();
        result.add(1);
        
        if(rowIndex==0) return result;
        
        List<Integer> prev = getRow(rowIndex-1);        
        for(int i=0; i<prev.size()-1;i++){
            result.add(prev.get(i)+prev.get(i+1));
        }
        result.add(1);
        
        return result;
    }

Problem solution in C++.

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        ++rowIndex;
        std::vector<int> res(rowIndex, 1);
        for(int i{1}; i < rowIndex; ++i) res[i] = (int64_t) res[i-1] * (rowIndex-i) / i;
        return res;
    }
};

Problem solution in C.

int* getRow(int rowIndex, int* returnSize) {
    int *result = malloc((rowIndex + 1) * sizeof(int));
    int i, j, new, old = 1;
    
    *returnSize = rowIndex + 1;
    
    for (i = 0; i <= rowIndex; i++) {
        result[i] = 1;
        for (j = 1; j < i; j++) {
            new = result[j];
            result[j] += old;
            old = new;
        }
    }
    
    return result;
}