LeetCode # Array # Easy #118. Pascal's Triangle

Title: Also known as "Yang Hui's Triangle" 

Given a non-negative integer numRows, generate the first numRows of Pascal's triangle.

In Pascal's triangle, each number is the sum of the two numbers directly above it.

Idea: use dynamic programming. First, the number of row 0 is 1; second, the first and last of each row are 1; finally, the other values ​​are the values ​​to the left of his previous row + the right of the previous row (bordering it).

How to do it: Think of the triangle as a big list, and store each row in the big list as a list of rows.

1  class Solution {
 2      public List<List<Integer>> generate( int numRows) {
 3          List<List<Integer>> triangle = new ArrayList<List<Integer>> ();//Store the entire triangle, where each row is also A list of integers.
4          if (numRows == 0 ){//When 0 rows are required, output an empty list;
 5              return triangle;
 6          }
 7          triangle.add( new ArrayList<> ());//The first row is always 1
 8          triangle.get(0).add(1 ); 
 9          for ( int numRow = 1; numRow < numRows; numRow ++ ){
 10             List<Integer> row = new ArrayList<> ();
 11              List<Integer> prevrow = triangle.get(numRow-1 );
 12              row.add(1 );//The first of each row=1
 13              for ( int j =1; j < numRow ; j++ ){
 14                  row.add(prevrow.get(j-1)+ prevrow.get(j));//The middle element is equal to the sum of the left and right of the previous row;
 15              }
 16              row.add(1 );//The last of each row=1;
 17              triangle.add(row);
 18          }
 19          return triangle;
 20      }  
 21 }