LeetCode(118)-pascal triangle(楊輝三角)

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]
]

具體生成算是：每一行的首個和結尾一個數字都是1，從第三行開始，中間的每個數字都是上一行的左右兩個數字之和

public class l118_pascaltriangle {


    public ArrayList<ArrayList<Integer>> generate(int numRows) {
        ArrayList<ArrayList<Integer>> res=new ArrayList<>(numRows);
        for(int i=0;i<numRows;i++){
            ArrayList<Integer> temp=new ArrayList<>(i);
            for(int j=0;j<=i;j++){
                if(j==0||j==i)//第一個和最後一個是1
                    temp.add(1);
                else {
                    temp.add(res.get(i-1).get(j-1)+res.get(i-1).get(j));

                }

            }
            res.add(temp);
        }
        return res;

    }

LeetCode(119)-pascal triangle 2
Given an index k, return the k th row of the Pascal’s triangle.

For example, given k = 3,
Return[1,3,3,1].

Note:
Could you optimize your algorithm to use only O(k) extra space?

public class l119_pascaltriangle2 {
    public ArrayList<Integer> getRow(int rowIndex) {
        ArrayList<ArrayList<Integer>> res=new ArrayList<>();
        for(int i=0;i<=rowIndex;i++){
            ArrayList<Integer> temp=new ArrayList<>();
            for(int j=0;j<=i;j++){
                if(j==0||j==i)
                    temp.add(1);
                else
                    temp.add(res.get(i-1).get(j-1)+res.get(i-1).get(j));

            }
            res.add(temp);
            if(i==rowIndex){
                return temp;
            }

        }
        return null;

    }