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

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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?

 

上一道題的延伸版，就是直接求出第k行的數，要求用o(k)的空間複雜度。

 

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;            }}

 

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