LeetCode:Pascal's Triangle I II

LeetCode: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]
]

分析：简单的模拟从第一层开始计算即可

class Solution {
public:
    vector<vector<int> > generate(int numRows) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        vector<vector<int> >res;
        if(numRows == 0)return res;
        res.push_back(vector<int>{1});
        if(numRows == 1)return res;
        vector<int>tmp;
        tmp.reserve(numRows);
        for(int i = 2; i <= numRows; i++)
        {
            tmp.clear();
            tmp.push_back(1);
            for(int j = 1; j < i-1; j++)
                tmp.push_back(res[i-2][j-1]+res[i-2][j]);
            tmp.push_back(1);
            res.push_back(tmp);
        }
        return res;
    }
};

 

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LeetCode: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?

分析：每次计算只和上一层有关系，因此只需要一个数组空间就可以                                                                                                            本文地址

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        vector<int> res(rowIndex+1,1);
        if(rowIndex == 0)return res;
        for(int i = 1; i <= rowIndex; i++)
        {
            int tmp = res[0];
            for(int j = 1; j <= i-1; j++)
            {
                int kk = res[j];
                res[j] = tmp+res[j];
                tmp = kk;
            }
        }
        return res;
    }
};

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