Pascal's Triangle & 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;
        vector<int> row(numRows, 1);
        if(numRows == 1)
            result.push_back(row);
        else if(numRows > 1)
        {
            result = generate(numRows - 1);    //使用递归的方式
            for(int i = 1; i < numRows - 1; i++)
                row[i] = result[numRows - 2][i - 1] + result[numRows - 2][i];
            result.push_back(row);
        }
        return result;
    }
};

 

Given an index k, return the k^th 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?

• 递归方法

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int> result(rowIndex + 1, 1);       //这里和上面I不一样的是，rowindex为0时，为第一行
        if(rowIndex == 0)
            return result;
        else
        {
            vector<int> tmp = getRow(rowIndex - 1);
            for(int i = 1; i < rowIndex; i++)
                result[i] = tmp[i - 1] + tmp[i];
            return result;
        }
    }
};

• 非递归

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