leetcode Pascal's Triangle

给定行号，输出如下所示Pascal Triangle（杨辉三角）

For example, given numRows = 5,
Return

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

思路，想到右上一个构建下一个，构成过程就是上一个的相邻元素相加，并且头尾为1.

class Solution {
public:
vector<int> fun116(vector<int> perm)
{
    vector<int> tmp;
    tmp.push_back(1);
    for (int i = 0; i < perm.size() - 1; i++)
    {
        tmp.push_back(perm[i] + perm[i + 1]);
    }
    tmp.push_back(1);
    return tmp;
}
vector<vector<int> > generate(int numRows)
{
    vector<vector<int> > ans;
    if (numRows == 0) return ans;
    vector<int> perm(1,1);
    while(numRows-- > 0)
    {
        ans.push_back(perm);
        perm = fun116(perm);
    }
    return ans;
}
};

　　也可以这样：

class Solution {
public:
    vector<vector<int> > generate(int numRows) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector<vector<int> > res;
        if(numRows == 0)
            return res;
        for(int i = 1; i <= numRows; i++)
        {
            vector<int> onelevel;
            onelevel.clear();
            onelevel.push_back(1);
            for(int j = 1; j < i; j++)
            {
                onelevel.push_back(res[i-2][j-1] + (j < i-1 ? res[i-2][j] : 0));
            }
            res.push_back(onelevel);
        }
        return res;
    }
};

 2015/03/31：

python：

class Solution:
    # @return a list of lists of integers
    def generate(self, numRows):
        mylist = [[1] for i in range(numRows)]
        for i in range(1, numRows):
            for j in range(1, i):
                mylist[i].append(mylist[i-1][j-1] + mylist[i-1][j])
            mylist[i].append(1)
        return mylist