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;
}