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;
}
}