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]
]

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/** * Return an array of arrays. * The sizes of the arrays are returned as *columnSizes array. * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free(). */ int** generate(int numRows, int** columnSizes) { if(numRows == 0) return NULL; int **result = (int **)malloc(numRows * sizeof(int *)); (*columnSizes) = (int *)malloc(numRows * sizeof(int)); int i, j; for (i = 0; i < numRows; i++) { (*columnSizes)[i] = i + 1; result[i] = (int *)malloc((*columnSizes)[i] * sizeof(int)); result[i][0] = 1; for (j = 1; j < i; j++) { result[i][j] = result[i - 1][j - 1] + result[i - 1][j]; } result[i][i] = 1; } return result; }

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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?

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class Solution { public: vector<int> getRow(int rowIndex) { vector<int> result; int n = rowIndex / 2; result.push_back(1); for(int i = 1; i <= n; i++){ long long next = (long(result[i - 1]) * long(rowIndex - i + 1)) / i; result.push_back(next); } if(rowIndex % 2 == 1){ int size = result.size(); for(int i = size - 1; i >= 0; i--){ result.push_back(result[i]); } }else{ int size = result.size(); for(int i = size - 2; i >= 0; i--){ result.push_back(result[i]); } } return result; } };

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