leetcode 119 Pascal's Triangle II ----- java

Given an index k, return the kth 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?

上一道题的延伸版，就是直接求出第k行的数，要求用o(k)的空间复杂度。

也是直接相加就可以了。

public class Solution {

    public List<Integer> getRow(int rowIndex) {

        List ans = new ArrayList<Integer>();

        if( rowIndex == 0){

            ans.add(1);

            return ans;

        }

        else if( rowIndex == 1){

            ans.add(1);

            ans.add(1);

            return ans;

        }    

        int[] result = new int[rowIndex+1];

        result[0] = 1;

        result[1] = 1;

        for( int i = 2;i<rowIndex+1;i++){

            int a = result[0];

            int b = result[1];

            result[i] = 1;

            for( int j = 1;j<i;j++){

                result[j] = a+b;

                a = b;

                b = result[j+1];

            }

        }

        for( int i = 0 ;i<rowIndex+1;i++)

            ans.add(result[i]);

        return ans;

    }

}

巴特西

leetcode 119 Pascal's Triangle II ----- java

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