Leetcode 119 Pascal ' s Triangle II (Pascal Triangle II) (vector, mathematical formula) (*)

translation

给定一个索引K，返回帕斯卡三角形的第K行。例如，给定K=3，返回[1,3,3,1]。注释：你可以改进你的算法只用O(k)的额外空间吗？

Original

returnof the Pascal‘s triangle.For3,Return [1,3,3,1touse only O(k) extra space?

Analysis

This question is actually to undertake the previous question, I also just finished writing.

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

The previous question is to return the complete Pascal triangle:

classSolution { Public: vector<vector<int>>GenerateintNumRows) { vector<vector<int>>Pascalif(NumRows <1)returnPascal vector<int>Root Root.push_back (1); Pascal.push_back (root);if(NumRows = =1)returnPascal Root.push_back (1); Pascal.push_back (root);if(NumRows = =2)returnPascalif(NumRows >2) { for(inti =2; i < numrows; ++i) { vector<int>Temp Temp.push_back (1); for(intj =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); }returnPascal }    }};

So I stole a lazy, since it is index K, then return it, but efficiency is ...

classSolution { Public: vector<int>GetRow (intRowIndex) {RowIndex + =1; vector<vector<int>>Pascalif(RowIndex <1)returnpascal[0]; vector<int>Root Root.push_back (1); Pascal.push_back (root);if(RowIndex = =1)returnpascal[0]; Root.push_back (1); Pascal.push_back (root);if(RowIndex = =2)returnpascal[1];if(RowIndex >2) { for(inti =2; i < RowIndex; ++i) { vector<int>Temp Temp.push_back (1); for(intj =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); }returnPascal[rowindex-1]; }    }};

A more efficient approach should be to have some kind of formula, to see what others are writing about ...

 vector<int>GetRow (intRowIndex) { vector<int>R R.resize (RowIndex +1); r[0] = R[rowindex] =1; for(Autoi =1; I < (r.size () +1) /2; ++i) {R[i] = r[rowindex-i] = (unsigned Long) R[i-1] * (unsigned Long) (Rowindex-i +1)/I; }returnR;}

Sure enough, the power of mathematics was revealed again. Let me write this formula in Markdown grammar ...

r i =r i 1 ?(INd ex?I+1)/I

Leetcode 119 Pascal ' s Triangle II (Pascal Triangle II) (vector, mathematical formula) (*)