PROBLEM DESCRIPTION
Given an integer A, equal to numRows, generate the first numRows of Pascal’s triangle.
Pascal’s triangle: To generate A[C] in row R, sum up A'[C] and A'[C-1] from the previous row R - 1.
SOLUTION
A[row][col] = A[row-1][col-1] + A[row-1][col]
This is pretty straight forward, the only thing to take care of is that the column can be out of bounds when looking into the previous row.
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public class Solution {
public ArrayList<ArrayList<Integer>> solve(int A) {
int n = A;
ArrayList<ArrayList<Integer>> triangle = new ArrayList<>();
if(n == 0) return triangle;
ArrayList<Integer> firstRow = new ArrayList<>();
firstRow.add(1);
triangle.add(firstRow);
for(int r=1; r<n; r++){
ArrayList<Integer> row = new ArrayList<>();
for(int c=0; c<=r; c++){
int nextVal = 0;
int previousRow = r-1;
// 1
// x y
// a b c
//the left column of x will be out of bounds, so handle that
if(c-1 >= 0)
nextVal += triangle.get(previousRow).get(c-1);
// 1
// x y
// a b c
//the column above the last element in each row will be out of bounds for the previous row, so handle that
if(c < triangle.get(previousRow).size())
nextVal += triangle.get(previousRow).get(c);
row.add(nextVal);
}
triangle.add(row);
}
return triangle;
}
}