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?

这个问题比较简单。

vector<int> getRow(int rowIndex) {
    vector<int> row, prev_row;

    row.push_back(1);

    for (int i = 1; i <= rowIndex; ++i) {
        prev_row = row;

        row.clear();
        row.push_back(1);
        for (int j = 1; j < prev_row.size(); j++)
            row.push_back(prev_row[j - 1] + prev_row[j]);
        row.push_back(1);
    }

    return row;
}

更简洁的做法：

vector<int> getRow(int rowIndex) {
    vector<int> row(rowIndex + 1, 0);

    row[0] = 1;
    for (int i = 1; i <= rowIndex; ++i) {
        for (int j = i + 1; j > 0; --j) {
            row[j] += row[j - 1];
        }
    }

    return row;
}

LeetCode[Array]: Pascal's Triangle II