leetcode:Pascal's Triangle II

vector<int> getRow(int rowIndex) {
        int *arr = new int[rowIndex+2];
        int *arr1 = new int[rowIndex+2];
        std::vector<int> v;
        if (rowIndex == 0)
        {
            v.push_back(1);
            return v;
        }
        arr[1] = 1;
        arr[2] = 1;
        int *src, *des;
        for (int i = 2; i <= rowIndex; ++i)
        {
            if ((i & 1)==0){
                src = arr;
                des = arr1;
            }
            else{
                src = arr1;
                des = arr;
            }
            des[1] = 1;
            des[2] = src[2] + 1;
            //cout << des[2] << endl;
            for (int j = 3; j < i+1; ++j)
            {
                des[j] = src[j - 1] + src[j];
            }
            des[i+1] = 1;
        }
        if (rowIndex & 1) src = arr;
        else src = arr1;
        for (int i = 1; i <= rowIndex+1; ++i)
        {
            v.push_back(src[i]);
        }
        delete[] arr;
        delete[] arr1;
        return v;
    }

最初的想法，通过两个一维数组迭代，因为当前行的值需要上一行的值，看完别人写的发现自己逗B了，其实可以发现，pascal三角形每行相比前行多一个数，那么可以通过一个数组来做，先将最后一个数右移一位，然后按公式算就行，代码如下

vector<int> getRow(int rowIndex) {
    std::vector<int> v;
    int i,j;
    int *a=new int[rowIndex+1];
    for (i = 0; i <= rowIndex; ++i)
    {
        a[0]=1;
        a[i]=1;
        for (j=i-1;j>0;--j)
        {
            a[j]=a[j-1]+a[j];
        }
    }
    for (i = 0; i <= rowIndex; ++i)
    {
        v.push_back(a[i]);
    }
    delete[] a;
    return v;
}