【leetcode】Pascal's Triangle I&II

Given numRows, generate the first numRows of Pascal's triangle.

For example, given numRows = 5,
Return

[
　　[1],
　 [1,1],
[1,2,1],
[1,3,3,1],
[1,4,6,4,1]
]

class Solution {
public:
vector<vector<int> > generate(int numRows) {

vector<vector<int>> result;
if(numRows==0) return result;
vector<int> tem;
tem.push_back(1);
result.push_back(tem);
if(numRows==1) return result;
tem.push_back(1);
result.push_back(tem);
if(numRows==2) return result;
for(int i=2;i<numRows;i++){
vector<int> solu(i+1,1);
for(int j=1;j<i;j++){
solu[j] = result[i-1][j-1]+result[i-1][j];
}
result.push_back(solu);
}
return result;
}

};

II:

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?

class Solution {
public:
vector<int> getRow(int rowIndex) {
vector<int> array;
for (int i = 0; i <= rowIndex; i++) {
for (int j = i - 1; j > 0; j--) {
array[j] = array[j - 1] + array[j];
}
array.push_back(1);

}
return array;
}
};