技術標籤：algorithm-design演算法leetcode

119.Pascal's Triangle II

Easy

Given an integerrowIndex, return therowIndexthrow of the Pascal's triangle.

Noticethat the row index starts from0.

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

Follow up:

Could you optimize your algorithm to use onlyO

Example 1:

Input: rowIndex = 3
Output: [1,3,3,1]

Example 2:

Input: rowIndex = 0
Output: [1]

Example 3:

Input: rowIndex = 1
Output: [1,1]

Constraints:

• 0 <=rowIndex <= 33

解法1：

純數學法。楊輝三角形每行就是
比如第3行(從0開始)，.
那麼，我們怎麼從呢？
根據可得
程式碼如下：
注意coeff中間可能會越界，所以要開long long。
時間複雜度O(n)，空間複雜度O(1)。

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int> res;
        long long coeff = 1;
        res.push_back(1);
        for (int i = 1; i <= rowIndex; i++) {
            coeff = coeff * (rowIndex - i + 1) / i;
            res.push_back((int)coeff) ;
        }
        return res;
    }
};
 

解法2:
迭代 。我用了兩個迴圈，而且還有一個輔助vector。時間複雜度O(n^2)，空間複雜度O(n)，效率不高。

```
#include <iostream>
#include <vector>

using namespace std;

vector<int> getRow(int rowIndex) {
if (rowIndex==0) return vector<int>(1,1);
if (rowIndex==1) return vector<int>(2,1);

vector<int> result(rowIndex+1, 1);

for (int i=2; i<=rowIndex; ++i) {
vector<int> prevResult=result;
for (int j=1; j<i; ++j) {
result[j]+=prevResult[j-1];
}
}

return result;
}

int main()
{
for (int k=0; k<=4; ++k) {
vector<int> a=getRow(k);
for (auto i:a) cout<<i<<" ";
cout<<endl;
}
return 0;
}