Given an integer numRows, return the first numRows of Pascal’s triangle.
In Pascal’s triangle, each number is the sum of the two numbers directly above it as shown:

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狀態(tài)轉(zhuǎn)移方程
triangle[i][j] = triangle[i-1][j-1] + triangle[i-1][j]

邊界條件
triangle[0][0] = triangle[i][0] = triangle[i][i] = 1

代碼

class Solution { public:vector<vector<int>> generate(int numRows) {vector<vector<int>> triangle;vector<int> temp;for(int i = 0; i < numRows; i++){temp.clear();temp.push_back(1);for(int j = 1; j < i; j++){temp.push_back(triangle[i-1][j-1] + triangle[i-1][j]);}if(i != 0)temp.push_back(1);triangle.push_back(temp);}return triangle;} };

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