[Leetcode]15. Pascal ' s triangle Yang Hui triangle

Given numrows, generate the first numrows of Pascal ' s triangle.

For example, given numrows = 5,
Return

[     1], [    1,2,1], [   1,3,3,1], [  1,4,6,4,1]

The Yang Hui triangle mainly has the following five properties:

1. The Yang Hui Triangle is made up of positive integers, the numbers are symmetrical, each line starts to grow larger by 1, then becomes smaller and goes back to 1.
2. The number of digits in the first row is one.
3. The first digit of the line is the number of combinations.
4. The number of the first row and the.
5. In addition to the leftmost and rightmost digits of each line, each number is equal to the sum of the two digits above its upper-left and upper-right (that is, the first digit of the line equals the sum of the first and the first digits of the first line). This is because there are combinatorial identities:.

Solution 1: Constructed directly according to the construction method of Yang Hui triangle.

classSolution { Public: Vector<vector<int>> Generate (intnumrows) {Vector< vector<int> >Res;  for(inti =0; i < numrows; i++) {vector<int> Row (i +1,1);  for(intj =1; J < I; J + +) Row[j]= Res[i-1][j-1] + res[i-1][j];        Res.push_back (row); }        returnRes; }};

Solution 2: The elements of each row of the Yang Hui triangle are the combined number C (m,n), where n represents the first few rows, m=0,1,..., N, using the formula C (m,n) =n!/m!/(N-M)!

classSolution { Public: Vector<vector<int>> Generate (intnumrows) {Vector< vector<int> >Res;  for(inti =0; i < numrows; i++) {vector<int> Row (i +1,1);  for(intj =0; J <= I; J + +) Row[j]= Fact (i)/fact (J)/fact (I-j);        Res.push_back (row); }        returnRes; }Private:    Long LongFactintN) {if(N <=1)            return 1; Long Longf =1, Res;  for(inti =2; I <= N; i++) {res= f *i; F=Res; }        returnRes; }};

Note that this does not pass the full test because the numrows is overflow when the calculation is large. Additional formula C (M,n) =c (m-1,n-1) +c (m,n-1) Considering the calculation of combinatorial numbers can be improved:

classSolution { Public: Vector<vector<int>> Generate (intnumrows) {Vector< vector<int> >Res;  for(inti =0; i < numrows; i++) {vector<int> row =GetRow (i);        Res.push_back (row); }        returnRes; }Private: Vector<int> GetRow (intRowIndex) {Vector<int>Row; if(RowIndex <0)             returnRow; Row.assign (RowIndex+1,0);  for(inti =0; I <= RowIndex; ++i) {if(i = =0) {row[0] =1; Continue; }             for(intj = RowIndex; J >=1; --j) Row[j]= Row[j] + row[j-1]; }        returnRow; }};

This method is inefficient because, in order to calculate the data for row row, all the data from line 1th to line row-1 is calculated first.

[Leetcode]15. Pascal ' s triangle Yang Hui triangle