118.杨辉三角

problems：
Given a non-negative integer numRows, generate the first numRows of Pascal’s triangle.

In Pascal’s triangle, each number is the sum of the two numbers directly above it.

Example:

Input: 5
Output:
[
[1],
[1,1],
[1,2,1],
[1,3,3,1],
[1,4,6,4,1]
]

solutions：
1.结果正确，但是会超时。

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    vector<vector<int>> generate(int numRows) {
        vector<vector<int>> vec;
        int n = numRows-1;
        if(numRows == 1)
        {
            vector<int> tip;
            tip.push_back(1);
            vec.push_back(tip);
        }
        else
        {
            vec = generate(n);
            vector<int> tip;
            tip.push_back(1);
            for(int i=0;i<n-1;i++)
           {
            int m=vec[n-1][i]+vec[n-1][i+1];
            tip.push_back(m);
           }
            tip.push_back(1);
            vec.push_back(tip);
        }
        return vec;
    }
};
int main()
{
    Solution so;
    int n;
    cin>>n;
    vector<vector<int>> vec;
    vec = so.generate(n);
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<=i;j++)
        {
            cout<<vec[i][j];
        }
        cout<<endl;
    }
    return 0;
}

2.直接构造法，可以通过。

class Solution {
public:
    vector<vector<int>> generate(int numRows) {
        vector<vector<int>> vec;
        if(numRows == 0) return vec;
        else vec.push_back({1});
        for(int i=1;i<numRows;i++)
        {
            vector<int> tip;
            for(int j=0;j<=i;j++)
            {
                if(j==0) tip.push_back(1);
                else
                if(j==i) tip.push_back(1);
                else
                    tip.push_back(vec[i-1][j-1]+vec[i-1][j]);
            }
            vec.push_back(tip);
        }
        return vec;
    }
};