LeetCode: Pascal's Triangle I && II

Title:

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]
]

递归：

class Solution {
public:
    vector<vector<int> > results;
    vector<vector<int>> generate(int numRows) {
        if (numRows < 1)
            return results;
        maker(numRows);
        return results;
    }
    
    void maker(int numRows){
        vector<int> cur;
        cur.push_back(1);
        if (numRows == 2){
            maker(numRows-1);
            cur.push_back(1);
        }
        else if (numRows > 2){
            maker(numRows-1);
            vector<int> pre = results[numRows-2];
            for (int i = 0 ; i < pre.size()-1; i++){
                cur.push_back(pre[i]+pre[i+1]);
            }
            cur.push_back(1);
        }
        results.push_back(cur);
    }
};

非递归

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

Title:

Given an index k, return the kth row of the Pascal's triangle.

For example, given k = 3,
Return [1,3,3,1].

思路：使用滚动数组

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