letecode [118] - Pascal's Triangle

Given a non-negative integer numRows, generate the first numRows of Pascal's triangle.

In Pascal's triangle, each number is the sum of the two numbers directly above it.

Example:

Input: 5
Output:
[
     [1],
    [1,1],
   [1,2,1],
  [1,3,3,1],
 [1,4,6,4,1]
]

 

题目大意：

　　输出前n行的杨辉三角，用二维vector表示。

理  解 ：

　　初始化前2行的结果。第3-n行，利用前一行的累加获得。

代码C++：

class Solution {
public:
    vector<vector<int>> generate(int numRows) {
        vector<vector<int>> vec;
        if(numRows==0) return vec;
        vector<int> rowVec;
        rowVec.push_back(1);
        vec.push_back(rowVec);
        if(numRows==1) return vec;
        rowVec.push_back(1);
        vec.push_back(rowVec);
        if(numRows==2) return vec;  
        int i=3;
        while(i<=numRows){
            rowVec.clear();
            rowVec.resize(i);
            rowVec[0] = 1;
            int j = 1;
            while(j<i-1){
                rowVec[j] = vec[i-2][j-1] + vec[i-2][j];
                ++j;
            }
            rowVec[i-1] = 1;
            vec.push_back(rowVec);
            ++i;
        }
        return vec;
    }
};

运行结果：

　　执行用时 : 0 ms　　内存消耗 : 8.8 MB