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; }
}

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