Given numRows, generate the first numRows of Pascal’s triangle.

For example, given numRows = 5,
Return

[
     [1],
    [1,1],
   [1,2,1],
  [1,3,3,1],
 [1,4,6,4,1]
]

public class Solution {
    public List<List<Integer>> generate(int numRows) {
        List<List<Integer>> pascalTriangle = new ArrayList<List<Integer>>();
        List<Integer> currLevel = new ArrayList<Integer>();
        currLevel.add(1);
        while(numRows>0){
            pascalTriangle.add(new ArrayList<Integer>(currLevel));
            currLevel = generateNextLevel(currLevel);
            numRows--;
        }
        return pascalTriangle;
    }
    
    public List<Integer> generateNextLevel(List<Integer> currLevel){
        List<Integer> nextLevel = new ArrayList<Integer>();
        if(currLevel.size()!=0){
            currLevel.add(0);
            currLevel.add(0,0);
            for(int runner=0; runner<currLevel.size()-1; runner++){
                nextLevel.add(currLevel.get(runner)+currLevel.get(runner+1));
            }
        }
        return nextLevel;
    }
}

Note: for each row, pretend there is always one 0 at each side of the row. So the triangle would look like following, so when calculating next level, just need to sum each adjacent integers in the current level.

[
     0[1]0,
    0[1,1]0,
   0[1,2,1]0,
  0[1,3,3,1]0,
 0[1,4,6,4,1]0
]

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