119. Pascal’s Triangle II

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:

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

From: LeetCode
Link: 119. Pascal’s Triangle II

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Solution:

Ideas:

1. Memory Allocation: The function uses malloc to allocate memory for the array that will store the rowIndex-th row. The size of the array is rowIndex + 1.

2. Initialization: The first element of each row is always 1, so row[0] = 1.

3. Calculating Row Elements:

• The outer loop runs from 1 to rowIndex to construct each row up to the desired rowIndex.
• Inside the outer loop, the last element of each row is set to 1.
• The inner loop updates the elements of the row in reverse order (from right to left) to avoid overwriting the values needed for the calculations. Each element row[j] is updated as the sum of row[j] and row[j-1].

4. Returning the Result: The function returns the pointer to the array containing the desired row.

Code:

/*** Note: The returned array must be malloced, assume caller calls free().*/
int* getRow(int rowIndex, int* returnSize) {*returnSize = rowIndex + 1;int* row = (int*)malloc((*returnSize) * sizeof(int));row[0] = 1;  // The first element is always 1for (int i = 1; i <= rowIndex; i++) {row[i] = 1;  // The last element in each row is always 1for (int j = i - 1; j > 0; j--) {row[j] = row[j] + row[j - 1];}}return row;
}