LeetCode#118 Pascal's Triangle

1、可以遍历,但折半显然好一点;
2、!!!一定注意“非负”;
3、对象的写法{{}}

class Solution {
public:
    vector<vector<int>> generate(int numRows) {

        if(numRows == 0) return {};
        if(numRows == 1) return {{1}};

        vector<vector<int>> triangle(numRows);

        triangle[0].resize(1);
        triangle[0][0] = 1;

        for(int i = 1; i < numRows; i++){
            triangle[i].resize(i+1);
            triangle[i][0] = 1;
            triangle[i][i] = 1;
            for(int j = 1; j < i; j++){
                if(j >= i/2+1){
                    triangle[i][j] = triangle[i][i - j];
                }else{
                    triangle[i][j] = triangle[i - 1][j - 1] + triangle[i - 1][j];
                }
            }
        }
        return triangle;

    }
};

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转载自blog.csdn.net/rpybd/article/details/81517371