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leetcode Pascal's Triangle

时间:2014-11-30 23:06:51      阅读:361      评论:0      收藏:0      [点我收藏+]

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;
    }
};

 

leetcode Pascal's Triangle

原文:http://www.cnblogs.com/higerzhang/p/4133855.html

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