LeetCode:Pascal's Triangle I II

LeetCode:Pascal's Triangle

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

For example, given numRows = 5,
Return

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

分析：简单的模拟从第一层开始计算即可

 class Solution {
 public:
     vector<vector<int> > generate(int numRows) {
         // IMPORTANT: Please reset any member data you declared, as
         // the same Solution instance will be reused for each test case.
         vector<vector<int> >res;
         if(numRows == )return res;
         res.push_back(vector<int>{});
         if(numRows == )return res;
         vector<int>tmp;
         tmp.reserve(numRows);
         for(int i = ; i <= numRows; i++)
         {
             tmp.clear();
             tmp.push_back();
             for(int j = ; j < i-; j++)
                 tmp.push_back(res[i-][j-]+res[i-][j]);
             tmp.push_back();
             res.push_back(tmp);
         }
         return res;
     }
 };

---

LeetCode:Pascal's Triangle II

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?

分析：每次计算只和上一层有关系，因此只需要一个数组空间就可以                                                                                                            本文地址

 class Solution {
 public:
     vector<int> getRow(int rowIndex) {
         // IMPORTANT: Please reset any member data you declared, as
         // the same Solution instance will be reused for each test case.
         vector<int> res(rowIndex+,);
         if(rowIndex == )return res;
         for(int i = ; i <= rowIndex; i++)
         {
             int tmp = res[];
             for(int j = ; j <= i-; j++)
             {
                 int kk = res[j];
                 res[j] = tmp+res[j];
                 tmp = kk;
             }
         }
         return res;
     }
 };

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