[Leetcode 36] 119 Pascal's Triangle II

Problem:

Given an index k, return the k^th 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?

Analysis:

The direct way to solve the problem is to use the formula: C(i, k) = k! / (i! * (k-i)!). But it will exceed the int number range if k is very large.

So we have to use the iterative way to construct the answer: to compute kth row, we first compute (k-1)th row and then use it to construct the desired row.

Code:

class Solution {
public:
 vector<int> getRow(int rowIndex) {
     vector<int> res;

     res.push_back(1);
     if (rowIndex == 0) return res;
     res.push_back(1);
     if (rowIndex == 1) return res;

     for (int i=1; i<rowIndex; i++) {
         vector<int> tmp;
         tmp.push_back(1);
         for (int j = 1; j <= i; j++) {
             tmp.push_back(res[j-1] + res[j]);
         }
         tmp.push_back(1);

         res = tmp;
     }

     return res;
 }

    
};

Attention:

Notice to add a new element into tmp, use push_back, not [].

The inner loop's max value is i not rowIndex.