Pascal’s Triangle – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given an integer numRows
, return the first numRows of Pascal’s triangle.
In Pascal’s triangle, each number is the sum of the two numbers directly above it as shown:
Example 1:
Input: numRows = 5 Output: [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]]
Example 2:
Input: numRows = 1 Output: [[1]]
Constraints:
1 <= numRows <= 30
C++ Pascal’s Triangle LeetCode Solution
class Solution {
public:
vector<vector<int> > generate(int numRows) {
vector<vector<int>> r(numRows);
for (int i = 0; i < numRows; i++) {
r[i].resize(i + 1);
r[i][0] = r[i][i] = 1;
for (int j = 1; j < i; j++)
r[i][j] = r[i - 1][j - 1] + r[i - 1][j];
}
return r;
}
};
Java Pascal’s Triangle LeetCode Solution
public class Solution {
public List<List<Integer>> generate(int numRows)
{
List<List<Integer>> allrows = new ArrayList<List<Integer>>();
ArrayList<Integer> row = new ArrayList<Integer>();
for(int i=0;i<numRows;i++)
{
row.add(0, 1);
for(int j=1;j<row.size()-1;j++)
row.set(j, row.get(j)+row.get(j+1));
allrows.add(new ArrayList<Integer>(row));
}
return allrows;
}
}
Python 3 Pascal’s Triangle LeetCode Solution
def generate(self, numRows):
res = [[1]]
for i in range(1, numRows):
res += [map(lambda x, y: x+y, res[-1] + [0], [0] + res[-1])]
return res[:numRows]
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