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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