LeetCode //C - 119. Pascal‘s Triangle II

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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:
在这里插入图片描述

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

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;
}