LeetCode 119 Pascal & amp; #39; s Triangle II (Pascal Triangle II) (vector, mathematical formula )(*)

LeetCode 119 Pascal & #39; s Triangle II (Pascal Triangle II) (vector, mathematical formula )(*)
Translation

Given an index K, return the K row of the Pascal triangle. For example, if K is 3, [1, 3, 3] is returned. Note: can you improve your algorithm to only use the extra space of O (k?

Original

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?

Analysis

This question is actually taken over by the previous question, and I just finished writing it.

LeetCode 118 Pascal's Triangle (Pascal Triangle) (vector)

The previous question is to return the complete Pascal triangle:

class Solution {public:    vector
  
   > generate(int numRows) {        vector
   
    > pascal;        if (numRows < 1) return pascal;        vector
    
      root;        root.push_back(1);        pascal.push_back(root);        if (numRows == 1)   return pascal;        root.push_back(1);        pascal.push_back(root);        if (numRows == 2)   return pascal;        if (numRows > 2) {            for (int i = 2; i < numRows; ++i) {                vector
     
       temp;                temp.push_back(1);                for (int j = 1; j < pascal[i - 1].size(); ++j) {                    temp.push_back(pascal[i - 1][j - 1] + pascal[i - 1][j]);                }                temp.push_back(1);                pascal.push_back(temp);            }            return pascal;        }    }};
     
    
   
  

So I am a little lazy. Since it is an index K, it is better to return it, but the efficiency is ......

class Solution {public:    vector
  
    getRow(int rowIndex) {        rowIndex += 1;        vector
   
    > pascal;        if (rowIndex < 1) return pascal[0];        vector
    
      root;        root.push_back(1);        pascal.push_back(root);        if (rowIndex == 1)  return pascal[0];        root.push_back(1);        pascal.push_back(root);        if (rowIndex == 2)  return pascal[1];        if (rowIndex > 2) {            for (int i = 2; i < rowIndex; ++i) {                vector
     
       temp;                temp.push_back(1);                for (int j = 1; j < pascal[i - 1].size(); ++j) {                    temp.push_back(pascal[i - 1][j - 1] + pascal[i - 1][j]);                }                temp.push_back(1);                pascal.push_back(temp);            }            return pascal[rowIndex - 1];        }    }};
     
    
   
  

A more efficient way should be to have a certain formula. Let's see what others have written ......

vector
  
    getRow(int rowIndex) {    vector
   
     r;    r.resize(rowIndex + 1);    r[0] = r[rowIndex] = 1;    for (auto i = 1; i < (r.size() + 1) / 2; ++i) {        r[i] = r[rowIndex - i] = (unsigned long)r[i - 1] * (unsigned long)(rowIndex - i + 1) / i;    }    return r;}
   
  

Sure enough, the power of mathematics appeared again. Let me use the Markdown syntax to write this formula ......

Ri = ri? 1? (Index? I + 1)/I