描述

Given an index k, return the kth row of the Pascal's triangle.

For example, given k = 3,

Return [1,3,3,1].

分析

先构造了一个杨辉三角，然后返回这个杨辉三角的最后一组值，没超时就万事大吉了...

感觉可以用递归写，但是嫌麻烦，放弃了。

代码如下：

class Solution {

public:

    vector<int> getRow(int rowIndex) {

        vector<int>ele; //store elements of current line

        vector<vector<int>>ret; //store all the lines and as a return variable

        if(rowIndex < 0)return ele;     //return an empty vector if rowIndex == 0

        int j = 0;  //initialize j

        for(int i = 0; i <= rowIndex; ++i){

            ele.push_back(1);   //push 1 before do any operators in a line

            if(i >= 1){

                j = 1;

                while(j < i){

                    ele.push_back(ret[i - 1][j - 1] + ret[i - 1][j]);

                    ++j;

                }

                ele.push_back(1);   //the last number in a line is also 1

            }

            ret.push_back(ele);

            ele.clear();    //each time we finish a line,clear this vector

        }

        return ret[rowIndex];

    }

};

巴特西

leetcode解题报告（24）：Pascal's TriangleII

描述

分析

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