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

1 class Solution { 2 public: 3     vector
     
      
        > generate(int numRows) { 4         // IMPORTANT: Please reset any member data you declared, as 5         // the same Solution instance will be reused for each test case. 6         vector
       
        
          >res; 7         if(numRows == 0)return res; 8         res.push_back(vector
         
          {
   1}); 9         if(numRows == 1)return res;10         vector
          
           tmp;11 tmp.reserve(numRows);12 for(int i = 2; i <= numRows; i++)13 {14 tmp.clear();15 tmp.push_back(1);16 for(int j = 1; j < i-1; j++)17 tmp.push_back(res[i-2][j-1]+res[i-2][j]);18 tmp.push_back(1);19 res.push_back(tmp);20 }21 return res;22 }23 };
          
         
        
       
      
     

 

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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                                                                                                            

1 class Solution { 2 public: 3     vector
     
       getRow(int rowIndex) { 4         // IMPORTANT: Please reset any member data you declared, as 5         // the same Solution instance will be reused for each test case. 6         vector
      
        res(rowIndex+1,1); 7         if(rowIndex == 0)return res; 8         for(int i = 1; i <= rowIndex; i++) 9         {10             int tmp = res[0];11             for(int j = 1; j <= i-1; j++)12             {13                 int kk = res[j];14                 res[j] = tmp+res[j];15                 tmp = kk;16             }17         }18         return res;19     }20 };
      
     

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