leetcode-pascal triangle I&&II

对于第2个pascal triangle,通过观察可以发现，其实只需要2个额外的变量来记录，于是就设了个tmp数组。

整体有点DP问题中的滚动数组的感觉。

 #include <vector>
 #include <iostream>
 using namespace std;

 class Solution {
 public:
     vector<vector<int> > generate(int numRows) {
         vector<vector<int>> res;
         if (numRows == ) return res;
         for (int i = ; i < numRows; i++)
         {
             vector<int> v; v.clear();
             for (int j = ; j <= i; j++)
             {
                 if (j ==  || j == i) v.push_back();
                 if (i> && j >  && j < i)
                 {
                     v.push_back(res[i - ][j] + res[i - ][j - ]);
                 }
             }
             res.push_back(v);
         }
         return res;///////////忘了返回了，一直找不出错来。
     }
     vector<int> getRow(int rowIndex) {
         vector<int> res(rowIndex+,);
         vector<int> tmp(,);
         //if (rowIndex == 0) return vector<int>(1,1);
         for (int i = ; i <= rowIndex; i++)
         {
             tmp[] = ; tmp[] = ;//别放错位置。之前放到内层的for里了。
             for (int j = ; j <= i; j++)
             {
                 if (j ==  || j == i)
                     res[j] = ;
                 if (j> && j < i)
                 {
                     if (j %  == )
                         tmp[] = res[j];
                     else if (j %  == )
                         tmp[] = res[j];
                     res[j] += tmp[-j%];
                 }
             }
         }
         return res;
     }
 };

 void printVV(vector<vector<int>> vv)
 {
     for (int i = ; i < vv.size(); i++)
     {
         for (int j = ; j < vv[i].size(); j++)
         {
             cout << vv[i][j] << " ";
         }
         cout << endl;
     }
 }
 int main()
 {
     Solution s;
     //vector<vector<int>> vv = s.generate(3);
     vector<int> v = s.getRow();
     for (int i = ; i < v.size(); i++)
     {
         cout << v[i] << " ";
     }
     //printVV(vv);
     return ;
 }