【leetcode】Pascal's Triangle I&II

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]
]

 

class Solution {
public:
    vector<vector<int> > generate(int numRows) {
        
        vector<vector<int>> result;
        if(numRows==0) return result;
    vector<int> tem;
    tem.push_back(1);
    result.push_back(tem);
    if(numRows==1) return result;
    tem.push_back(1);
    result.push_back(tem);
    if(numRows==2) return result;
    for(int i=2;i<numRows;i++){
        vector<int> solu(i+1,1);
        for(int j=1;j<i;j++){
            solu[j] = result[i-1][j-1]+result[i-1][j];
        }
        result.push_back(solu);
    }
    return result;
    }
    
};

 

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?

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int> array;
        for (int i = 0; i <= rowIndex; i++) {
            for (int j = i - 1; j > 0; j--) {
                array[j] = array[j - 1] + array[j];
            }
            array.push_back(1);
            
        }
        return array;
    }
};