[leetcode] 3. Pascal's Triangle

第三道还是帕斯卡三角，这个是要求正常输出，题目如下：

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]
]

那么这个就很简单了，按照帕斯卡三角也是杨辉三角的定义来就可以了。我的代码如下：

vector<vector<int>> generate(int numRows)
{
	vector<vector<int>> Pascal;
	vector<int> tmp;

	if (numRows == 0)
	{
		return Pascal;
	}

	tmp.push_back(1);
	Pascal.push_back(tmp);

	for (int i = 1; i < numRows; i++)
	{
		tmp.clear();

		tmp.push_back(1);
		for (int j = 1; j < i; j++)
		{
			tmp.push_back(Pascal[i - 1][j - 1] + Pascal[i - 1][j]);
		}
		tmp.push_back(1);
		Pascal.push_back(tmp);
	}

	return Pascal;
}

这个跟上一题有一个区别是在这里认为帕斯卡三角的第0层应该返回[]，而上一题中输入numRows==0时，返回的是[1]，这个在最开始加一个判断就可以了没必要纠结。

[leetcode] 3. Pascal's Triangle

时间： 2024-10-12 15:56:20

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