leetcode 119

119. Pascal's Triangle 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?

输出Pascal三角形的第k行（从0开始）；空间复杂度为O(k)；

从第0行开始生成，每次利用已经生成的行，在原有空间上生成下一行，直到生成所要求的行。

代码入下：

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int>  pascal;
        pascal.push_back(1);
        int m = 1;
        for(int i = 0; i <= rowIndex; i++)
        {
            if(rowIndex == 0)
            {
                return pascal;
            }
            for(int j = 1; j < i+1; j++)
            {
                if(j == i)
                {
                    pascal.push_back(1);
                    break;
                }
                int n = pascal[j];
                pascal[j] = m + n;
                m = n;
            }
        }
       return pascal;
    }
};