(leetcode题解)Pascal's Triangle

Pascal‘s Triangle 

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]
]

题意实现一个杨辉三角。

这道题只要注意了边界条件应该很好实现出来,C++实现如下

    vector<vector<int>> generate(int numRows) {
        vector<vector<int>> res;
        res.resize(numRows);
        for(int i=0;i<numRows;i++)
        {
            res[i].push_back(1);
            for(int j=1;j<i;j++)
            {
                res[i].push_back(res[i-1][j-1]+res[i-1][j]);
            }
            if(i!=0)
                res[i].push_back(1);
        }
        return res;
    }

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?

题意是给一个k值,返回杨辉三角的第k+1行,要求空间复杂度是O(k)。

这道题的其实本质与上一道一样求杨辉三角,申请一个空间保存上一行的值即可,C++实现如下:

vector<int> getRow(int rowIndex) {
        vector<int> res,temp;
        res.resize(rowIndex+1);
        for(int i=0;i<rowIndex+1;i++)
        {
            res[0]=1;
            for(int j=1;j<i;j++)
            {
                res[j]=temp[j-1]+temp[j];
            }
            if(i!=0)
                res[i]=1;
            temp=res;
        }
        return res;
    }

(leetcode题解)Pascal's Triangle

时间: 2024-10-04 02:39:34

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