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?
public class Solution {
public List<Integer> getRow(int rowIndex) {
if(rowIndex<0){
return null;
}
List<List<Integer>> list=new ArrayList<>();
if(rowIndex>=0){
List<Integer> l=new ArrayList<>();
l.add(1);
list.add(l);
}
if(rowIndex>=1){
List<Integer> l=new ArrayList<>();
l.add(1);
l.add(1);
list.add(l);
}
if(rowIndex>=2){
for(int i=2;i<=rowIndex;i++){
List<Integer>l=new ArrayList<>();
List<Integer>prev=list.get(i-1);
l.add(1);
for(int j=1;j<=i-1;j++){
l.add(prev.get(j-1)+prev.get(j));
}
l.add(1);
list.add(l);
}
}
return list.get(rowIndex);
}
}
Others' Solution
public List<Integer> getRow(int rowIndex) {
List<Integer> list = new ArrayList<Integer>();
if (rowIndex < 0)
return list;
for (int i = 0; i < rowIndex + 1; i++) {
list.add(0, 1);
for (int j = 1; j < list.size() - 1; j++) {
list.set(j, list.get(j) + list.get(j + 1));
}
}
return list;
}