Given an index k, return the kth 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?
class Solution {
public:
vector<int> getRow(int rowIndex) {
vector<int> vet1 = {1};
vector<int> vet2 = {1,1};
if (rowIndex == 0)
return vet1;
if (rowIndex == 1)
return vet2;
for (int i = 2; i <= rowIndex; i++)
{
if (i % 2 == 0)
{
vet1.clear();
vet1.push_back(1);
for (int i = 0; i < vet2.size() - 1; i++)
{
vet1.push_back(vet2[i] + vet2[i + 1]);
}
vet1.push_back(1);
}
else
{
vet2.clear();
vet2.push_back(1);
for (int i = 0; i < vet1.size() - 1; i++)
{
vet2.push_back(vet1[i] + vet1[i + 1]);
}
vet2.push_back(1);
}
}
if (rowIndex % 2 == 0)
{
return vet1;
}
else
return vet2;
}
};
别人的：
class Solution {
public:
vector<int> getRow(int rowIndex) {
vector<int> res(rowIndex + 1, 0); //
res[0] = 1;
for (int i = 1; i < rowIndex + 1; ++i)
{
for (int j = i; j > 0; --j)
res[j] += res[j - 1];
}
return res;
}
};
119. Pascal's Triangle II（LeetCode）
原文：http://www.cnblogs.com/wujufengyun/p/6922830.html